Problem: The king of Berland organizes a ball! n pair are invited to the ball, they are numbered from 1 to n. Each pair consists of one man and one woman. Each dancer (either man or woman) has a monochrome costume. The color of each costume is represented by an integer from 1 to k, inclusive.

Let b_i be the color of the man's costume and g_i be the color of the woman's costume in the i-th pair. You have to choose a color for each dancer's costume (i.e. values b_1, b_2, ..., b_n and g_1, g_2, ... g_n) in such a way that:

  1. for every i: b_i and g_i are integers between 1 and k, inclusive; 
  2. there are no two completely identical pairs, i.e. no two indices i, j (i ≠ j) such that b_i = b_j and g_i = g_j at the same time; 
  3. there is no pair such that the color of the man's costume is the same as the color of the woman's costume in this pair, i.e. b_i ≠ g_i for every i; 
  4. for each two consecutive (adjacent) pairs both man's costume colors and woman's costume colors differ, i.e. for every i from 1 to n-1 the conditions b_i ≠ b_{i + 1} and g_i ≠ g_{i + 1} hold. 



Let's take a look at the examples of bad and good color choosing (for n=4 and k=3, man is the first in a pair and woman is the second):

Bad color choosing: 

  * (1, 2), (2, 3), (3, 2), (1, 2) — contradiction with the second rule (there are equal pairs); 
  * (2, 3), (1, 1), (3, 2), (1, 3) — contradiction with the third rule (there is a pair with costumes of the same color); 
  * (1, 2), (2, 3), (1, 3), (2, 1) — contradiction with the fourth rule (there are two consecutive pairs such that colors of costumes of men/women are the same). 



Good color choosing: 

  * (1, 2), (2, 1), (1, 3), (3, 1); 
  * (1, 2), (3, 1), (2, 3), (3, 2); 
  * (3, 1), (1, 2), (2, 3), (3, 2). 



You have to find any suitable color choosing or say that no suitable choosing exists.

Input

The only line of the input contains two integers n and k (2 ≤ n, k ≤ 2 ⋅ 10^5) — the number of pairs and the number of colors.

Output

If it is impossible to find any suitable colors choosing, print "NO".

Otherwise print "YES" and then the colors of the costumes of pairs in the next n lines. The i-th line should contain two integers b_i and g_i — colors of costumes of man and woman in the i-th pair, respectively.

You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.

Examples

Input


4 3


Output


YES
3 1
1 3
3 2
2 3


Input


10 4


Output


YES
2 1
1 3
4 2
3 4
4 3
3 2
2 4
4 1
1 4
3 1


Input


13 4


Output


NO

Solution (in JAVA):

Solution:
import java.util.*;
import java.lang.*;
import java.io.*;
import java.awt.*;

// U KNOW THAT IF THIS DAY WILL BE URS THEN NO ONE CAN DEFEAT U HERE................
//JUst keep faith in ur strengths .................................................. 


// ASCII = 48 + i ;// 2^28 = 268,435,456  > 2* 10^8 // log 10 base 2 = 3.3219 
// odd:: (x^2+1)/2 , (x^2-1)/2 ; x>=3// even:: (x^2/4)+1 ,(x^2/4)-1  x >=4 
// FOR ANY ODD NO N : N,N-1,N-2
//ALL ARE PAIRWISE COPRIME 
//THEIR COMBINED LCM IS PRODUCT OF ALL THESE NOS

// two consecutive odds are always coprime to each other
// two consecutive even have always gcd  = 2 ;

// Rectangle r = new Rectangle(int x , int y,int widht,int height) 
//Creates a rect. with bottom left cordinates as (x, y) and top right as ((x+width),(y+height))

//BY DEFAULT Priority Queue is MIN in nature in java
//to use as max , just push with negative sign and change sign after removal 

// We can make a sieve of max size 1e7 .(no time or space issue) 
// In 1e7 starting nos we have about 66*1e4 prime nos 

 public class Main
{
     
    // static int[] arr = new int[100002] ; 
    // static int[] dp = new int[100002] ;  
    
     static PrintWriter out;
    
	static class FastReader{
		BufferedReader br;
		StringTokenizer st;
		public FastReader(){
			br=new BufferedReader(new InputStreamReader(System.in));
			out=new PrintWriter(System.out);
		}
		String next(){
			while(st==null || !st.hasMoreElements()){
				try{
					st= new StringTokenizer(br.readLine());
				}
				catch (IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt(){
			return Integer.parseInt(next());
		}
		long nextLong(){
			return Long.parseLong(next());
		}
		double nextDouble(){
			return Double.parseDouble(next());
		}
		String nextLine(){
			String str = "";
			try{
				str=br.readLine();
			}
			catch(IOException e){
				e.printStackTrace();
			}
			return str;
		}
	}
	


////////////////////////////////////////////////////////////////////////////////////
 public static int countDigit(long n) 
    { 
        return (int)Math.floor(Math.log10(n) + 1); 
    } 

///////////////////////////////////////////////////////////////////////////////////////// 
 
 public static int sumOfDigits(long n)
 {
  
  if( n< 0)return -1 ;
  
  int sum = 0;
  
  while( n > 0)
  {
      sum = sum + (int)( n %10) ;
      
      n /= 10 ;
  }
     
  return sum ;  
 
 
 
 }
 
 //////////////////////////////////////////////////////////////////////////////////////////////////

public static long arraySum(int[] arr , int start , int end)
{
    long ans = 0 ;
    
    for(int i = start ; i <= end  ; i++)ans += arr[i] ;
    
    return ans  ;
}

/////////////////////////////////////////////////////////////////////////////////



////////////////////////////////////////////////////////////////////////////////

public static void swapArray(int[] arr , int start , int end)
{
    while(start < end)
    {
        int temp = arr[start] ;
        arr[start] = arr[end];
        arr[end] = temp;
        start++ ;end-- ;
    }
}


//////////////////////////////////////////////////////////////////////////////////

static long factorial(long a)
{
    if(a== 0L || a==1L)return 1L ;
    
    return a*factorial(a-1L) ;
}

///////////////////////////////////////////////////////////////////////////////


public static int[][] rotate(int[][] input){

int n =input.length;
int m = input[0].length ;
int[][] output = new int [m][n];

for (int i=0; i<n; i++)
	for (int j=0;j<m; j++)
		output [j][n-1-i] = input[i][j];
return output;
}
///////////////////////////////////////////////////////////////////////////////////////////////////////


/////////////////////////////////////////// ////////////////////////////////////////////////   

public static boolean isPowerOfTwo(long n) 
{ 
    if(n==0) 
    return false; 
  
if(((n ) & (n-1)) == 0 ) return true ;
else return false  ;

} 

/////////////////////////////////////////////////////////////////////////////////////

 
/////////////////////////////////////////////////////////////////////////////////// 

public static String reverse(String input)
{
  StringBuilder str  = new StringBuilder("") ;
   
    for(int i =input.length()-1 ; i >= 0  ; i-- )
    {
        str.append(input.charAt(i));
    }
    
return str.toString() ;
}
///////////////////////////////////////////////////////////////////////////////////////////



////////////////////////////////////////////////////////////////////////////////////////////////////
public static boolean isPossibleTriangle(int a  ,int b , int c)
{
      if( a + b > c && c+b > a && a +c > b)return true  ;
      else return false  ;
}


////////////////////////////////////////////////////////////////////////////////////////////
static long xnor(long num1, long num2) {
		if (num1 < num2) {
			long temp = num1;
			num1 = num2;
			num2 = temp;
		}
		num1 = togglebit(num1);
		return num1 ^ num2;
	}

	static long togglebit(long n) {
		if (n == 0)
			return 1;
		long i = n;
		n |= n >> 1;
		n |= n >> 2;
		n |= n >> 4;
		n |= n >> 8;
		n |= n >> 16;
		return i ^ n;
	}

///////////////////////////////////////////////////////////////////////////////////////////////

public static int xorOfFirstN(int n)
{
 
 
 if( n % 4 ==0)return n ;
 
 else if( n % 4 == 1)return 1 ;
 
 else if( n % 4 == 2)return n+1 ;
 
 else return 0 ;
 
    
}
//////////////////////////////////////////////////////////////////////////////////////////////

public static int gcd(int a, int b )
{

if(b==0)return a ;

else return gcd(b,a%b) ; 

}


public static long gcd(long a, long b )
{

if(b==0)return a ;

else return gcd(b,a%b) ; 

}

////////////////////////////////////////////////////////////////////////////////////

public static int lcm(int a, int b ,int c , int d )
{

int temp = lcm(a,b , c) ;


 
 int ans = lcm(temp ,d ) ;

return ans  ;


}

///////////////////////////////////////////////////////////////////////////////////////////

public static int lcm(int a, int b ,int c )
{

int temp = lcm(a,b) ;

int ans =  lcm(temp ,c) ;

return ans  ;


}

////////////////////////////////////////////////////////////////////////////////////////
    
public static int lcm(int a , int b )
{

int gc = gcd(a,b);

return (a/gc)*b ;
}


public static long lcm(long a , long b )
{

long gc = gcd(a,b);

return (a/gc)*b;
}


///////////////////////////////////////////////////////////////////////////////////////////
static boolean isPrime(long n)
{
      if(n==1)
      {
            return false  ;
      }
      
      boolean ans =  true  ;
      
      for(long i = 2L; i*i <= n ;i++)
      {
            if(n% i ==0)
            {
                  ans = false  ;break ;
            }
      }
      
      
      return ans  ;
} 

static boolean isPrime(int n)
{
      if(n==1)
      {
            return false  ;
      }
      
      boolean ans =  true  ;
      
      for(int i = 2; i*i <= n ;i++)
      {
            if(n% i ==0)
            {
                  ans = false  ;break ;
            }
      }
      
      
      return ans  ;
}      


///////////////////////////////////////////////////////////////////////////

static int sieve =  1000000 ;

 
static boolean[] prime =  new boolean[sieve + 1] ;

public static void sieveOfEratosthenes() 
    { 
        // FALSE == prime
        
        // TRUE ==  COMPOSITE
        
        // FALSE== 1
       
        
        // time complexity = 0(NlogLogN)== o(N)
        
        // gives prime nos bw 1 to N
        
        for(int i = 4; i<= sieve ; i++)
        {
            prime[i] = true  ;
            i++ ;
        }
        
        for(int p = 3; p*p <= sieve; p++) 
        { 
           
            if(prime[p] == false) 
            { 
                
                for(int i = p*p; i <= sieve; i += p) 
                    prime[i] = true; 
            } 
            
            p++ ;
        } 
          
       
       
    
    } 
 
///////////////////////////////////////////////////////////////////////////////////


public static void sortD(int[] arr , int s  , int e)
{
      sort(arr ,s , e) ;
      
      int i =s ; int j = e  ;
      
      while( i < j)
      {
            int temp = arr[i] ;
            arr[i] =arr[j] ;
            arr[j] = temp ;
            i++ ; j-- ;
      }
      
      
      
      return ;
}


/////////////////////////////////////////////////////////////////////////////////////////

public static long countSubarraysSumToK(long[] arr ,long sum )
    {
      HashMap<Long,Long> map = new HashMap<>() ;
       
      int n = arr.length ;
       
      long prefixsum = 0 ;
       
      long count = 0L ;
      for(int i  = 0; i < n ; i++)
      {
          prefixsum  = prefixsum +  arr[i] ;
           
          if(sum == prefixsum)count = count+1 ;
           
          if(map.containsKey(prefixsum -sum))
          {
              count = count + map.get(prefixsum -sum) ;
          }
          
          
          if(map.containsKey(prefixsum ))
          {
              map.put(prefixsum , map.get(prefixsum) +1 );
          }
          
          else{
              map.put(prefixsum , 1L );
          }
           
           
      }
        
        
        
      return count  ;  
        
    }

///////////////////////////////////////////////////////////////////////////////////////////////


// KMP ALGORITHM : TIME COMPL:O(N+M) 
// FINDS THE OCCURENCES OF PATTERN AS A  SUBSTRING IN STRING
//RETURN THE ARRAYLIST OF INDEXES 
// IF SIZE OF LIST IS ZERO MEANS PATTERN IS NOT PRESENT IN STRING


public static ArrayList<Integer> kmpAlgorithm(String str , String pat)
     {
        ArrayList<Integer> list =new ArrayList<>();
        
        int n = str.length() ;
        int m = pat.length() ;
        
        String q = pat + "#" + str ;
        
        int[] lps  =new int[n+m+1] ;
        
         longestPefixSuffix(lps, q,(n+m+1)) ;
         
         
         for(int i =m+1 ; i < (n+m+1) ; i++ )
         {
             if(lps[i] == m)
             {
                 list.add(i-2*m) ;
             }
         }
        
        return list ; 
         
        
    }
    

public static void longestPefixSuffix(int[] lps ,String str , int n)
    {
        lps[0] =  0 ;
        
        for(int i = 1  ; i<= n-1; i++)
        {
          int l = lps[i-1] ;
           
          while( l > 0 && str.charAt(i) != str.charAt(l))
          {
              l = lps[l-1] ;
          }
           
          if(str.charAt(i) == str.charAt(l))
          {
              l++ ;
          }
            
           
          lps[i] = l ; 
        }
        
    }
    
     

///////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////

 // CALCULATE TOTIENT Fn FOR ALL VALUES FROM 1 TO n
    // TOTIENT(N) = count of nos less than n and grater than 1 whose gcd with n is 1 
    // or n and the no  will be coprime in nature
    //time : O(n*(log(logn)))
    
    public static void eulerTotientFunction(int[] arr ,int n )
    {
      
      for(int i = 1; i <= n  ;i++)arr[i] =i  ;
      
      
      for(int i= 2 ; i<= n ;i++)
      {
          if(arr[i] == i)
          {
              arr[i] =i-1 ;
              
              for(int j =2*i ; j<= n  ; j+= i )
              {
                  arr[j] = (arr[j]*(i-1))/i ;
              }
              
          }
      }
        
      return  ;  
        
    }
	
/////////////////////////////////////////////////////////////////////////////////////////////
public static long nCr(int n,int k)
{
    long ans=1L;
    k=k>n-k?n-k:k;
    int j=1;
    for(;j<=k;j++,n--)
    {
        if(n%j==0)
        {
            ans*=n/j;
        }else
        if(ans%j==0)
        {
            ans=ans/j*n;
        }else
        {
            ans=(ans*n)/j;
        }
    }
    return ans;
}

///////////////////////////////////////////////////////////////////////////////////////////

public static ArrayList<Integer> allFactors(int n)
{   
      ArrayList<Integer> list = new ArrayList<>() ;
      
    for(int i = 1; i*i <= n ;i++)
    {
          if( n % i == 0)
          {
              if(i*i == n)
              {
                    list.add(i) ;
              }
              else{
                    list.add(i) ;
                    list.add(n/i) ;
                    
              }
          }
    }
      
     return list ; 
      
      
}


public static ArrayList<Long> allFactors(long n)
{   
      ArrayList<Long> list = new ArrayList<>() ;
      
    for(long i = 1L; i*i <= n ;i++)
    {
          if( n % i == 0)
          {
              if(i*i == n)
              {
                    list.add(i) ;
              }
              else{
                    list.add(i) ;
                    list.add(n/i) ;
                    
              }
          }
    }
      
     return list ; 
      
      
}
////////////////////////////////////////////////////////////////////////////////////////////////////

  static final int MAXN = 1000001; 
       
    
    static int spf[] = new int[MAXN]; 
   
    static void sieve() 
    { 
        spf[1] = 1; 
        for (int i=2; i<MAXN; i++) 
       
           
            spf[i] = i; 
       
       
        for (int i=4; i<MAXN; i+=2) 
            spf[i] = 2; 
       
        for (int i=3; i*i<MAXN; i++) 
        { 
           
            if (spf[i] == i) 
            { 
               
                for (int j=i*i; j<MAXN; j+=i) 
       
                    if (spf[j]==j) 
                        spf[j] = i; 
            } 
        } 
    } 
       

    static ArrayList<Integer> getPrimeFactorization(int x) 
    { 
        ArrayList<Integer> ret = new ArrayList<Integer>(); 
        while (x != 1) 
        { 
            ret.add(spf[x]); 
            x = x / spf[x]; 
        } 
        return ret; 
    } 
       
 //////////////////////////////////////////////////////////////////////////////////////////////////
 //////////////////////////////////////////////////////////////////////////////////////////////////
   
public static void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        int L[] = new int[n1];
        int R[] = new int[n2];
 
       //Copy data to temp arrays
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
  public static void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

public static void sort(long arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }


public static void merge(long arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        long L[] = new long[n1];
        long R[] = new long[n2];
 
        //Copy data to temp arrays
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 

 /////////////////////////////////////////////////////////////////////////////////////////
 
	public static long knapsack(int[] weight,long value[],int maxWeight){

        
        int n=  value.length ;
        
	
	//dp[i] stores the profit with KnapSack capacity "i" 
  long []dp = new long[maxWeight+1]; 
      
    //initially profit with 0 to W KnapSack capacity is 0 
    Arrays.fill(dp, 0); 
  
    // iterate through all items 
    for(int i=0; i < n; i++)  
      
        //traverse dp array from right to left 
        for(int j = maxWeight; j >= weight[i]; j--) 
            dp[j] = Math.max(dp[j] , value[i] + dp[j - weight[i]]); 
              
    /*above line finds out maximum of dp[j](excluding ith element value) 
    and val[i] + dp[j-wt[i]] (including ith element value and the 
    profit with "KnapSack capacity - ith element weight") */
    return dp[maxWeight]; 
	}

///////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////


// to return max sum of any subarray in given array
public static long kadanesAlgorithm(long[] arr)
{
    
    if(arr.length == 0)return  0 ;
    
      long[] dp = new long[arr.length] ;
      
      dp[0] = arr[0] ;
      long max =  dp[0] ;
      
      
      for(int i = 1; i <  arr.length ; i++)
      {
            if(dp[i-1] > 0)
            {
                  dp[i] = dp[i-1] + arr[i] ;
            }
            else{
                  dp[i] = arr[i] ;
            }
            
            if(dp[i] >  max)max = dp[i] ;
            
      }
      
      return max  ;
      
}
/////////////////////////////////////////////////////////////////////////////////////////////
public static long kadanesAlgorithm(int[] arr)
{
     if(arr.length == 0)return  0 ;
    
      long[] dp = new long[arr.length] ;
      
      dp[0] = arr[0] ;
      long max =  dp[0] ;
      
      
      for(int i = 1; i <  arr.length ; i++)
      {
            if(dp[i-1] > 0)
            {
                  dp[i] = dp[i-1] + arr[i] ;
            }
            else{
                  dp[i] = arr[i] ;
            }
            
            if(dp[i] >  max)max = dp[i] ;
            
      }
      
      return max  ;
      
}

      
///////////////////////////////////////////////////////////////////////////////////////



/////////////////////////////////////////////////////////////////////////////////////////



//////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////

//TO GENERATE ALL(DUPLICATE ALSO EXIST) PERMUTATIONS OF A STRING


// JUST CALL generatePermutation( str,  start,  end) start :inclusive ,end : exclusive 

//Function for swapping the characters at position I with character at position j  
    public static String swapString(String a, int i, int j) {  
        char[] b =a.toCharArray();  
        char ch;  
        ch = b[i];  
        b[i] = b[j];  
        b[j] = ch;  
        return String.valueOf(b);  
    }  
      
//Function for generating different permutations of the string  
    public static void generatePermutation(String str, int start, int end)  
    {  
        //Prints the permutations  
        if (start == end-1)  
            System.out.println(str);  
        else  
        {  
            for (int i = start; i < end; i++)  
            {  
                //Swapping the string by fixing a character  
                str = swapString(str,start,i);  
                //Recursively calling function generatePermutation() for rest of the characters   
                generatePermutation(str,start+1,end);  
                //Backtracking and swapping the characters again.  
                str = swapString(str,start,i);  
            }  
        }  
    }  



////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////

public static long factMod(long n, long mod) {
    if (n <= 1) return 1;
    long ans = 1;
    for (int i = 1; i <= n; i++) {
      ans = (ans * i) % mod;
    }
    return ans;
  }


/////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////

public static long power(int a  ,int b)
    {
        //time comp : o(logn) 
        
        long x = (long)(a) ;
         long n = (long)(b) ;
        
        if(n==0)return 1 ;
        if(n==1)return x;
        
        long ans =1L  ;
       
      while(n>0)
      {
          if(n % 2 ==1)
          {
              ans = ans *x ;
          }
           
          n = n/2L ;
           
          x =  x*x ;
           
      }
       
      return ans ;
    }
    
    public static long power(long a  ,long b)
    {
        //time comp : o(logn) 
        
        long x = (a) ;
         long n = (b) ;
        
        if(n==0)return 1L ;
        if(n==1)return x;
        
        long ans =1L  ;
       
      while(n>0)
      {
          if(n % 2 ==1)
          {
              ans = ans *x ;
          }
           
          n = n/2L ;
           
          x =  x*x ;
           
      }
       
      return ans ;
    }

    
    
    

////////////////////////////////////////////////////////////////////////////////////////////////////
public static long powerMod(long x, long n, long mod) {
    //time comp : o(logn)
    
    if(n==0)return 1L ;
        if(n==1)return x;
        
    
    long ans = 1;
    while (n > 0) {
      if (n % 2 == 1) ans = (ans * x) % mod;
      x = (x * x) % mod;
      n /= 2;
    }
    return ans;
  }
 
//////////////////////////////////////////////////////////////////////////////////

/////////////////////////////////////////////////////////////////////////////////

/*
lowerBound - finds largest element equal or less than value paased
upperBound - finds smallest element equal or more than value passed

if not present return -1;

*/

public static long lowerBound(long[] arr,long k)
	{
		long ans=-1;
		
		int start=0;
		int end=arr.length-1;
		
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]<=k)
			{
				ans=arr[mid];
				start=mid+1;
			}
			else
			{
				end=mid-1;
			}
			
		}
		
		return ans;
		
	}
	
	public static int lowerBound(int[] arr,int k)
	{
		int ans=-1;
		
		int start=0;
		int end=arr.length-1;
		
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]<=k)
			{
				ans=arr[mid];
				start=mid+1;
			}
			else
			{
				end=mid-1;
			}
			
		}
		
		return ans;
		
	}
	
	
	public static long upperBound(long[] arr,long k)
	{
		long ans=-1;
		
		int start=0;
		int end=arr.length-1;
		
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]>=k)
			{
				ans=arr[mid];
				end=mid-1;
			}
			else
			{
				start=mid+1;
			}
			
		}
		
		return ans;
	}
	
	
	public static int upperBound(int[] arr,int k)
	{
		int ans=-1;
		
		int start=0;
		int end=arr.length-1;
		
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]>=k)
			{
				ans=arr[mid];
				end=mid-1;
			}
			else
			{
				start=mid+1;
			}
			
		}
		
		return ans;
	}
	

//////////////////////////////////////////////////////////////////////////////////////////

public static void printArray(int[] arr , int si ,int ei)
{
    for(int i = si  ; i  <= ei ; i++)
    {
        out.print(arr[i] +" ") ;
    }
    
}

public static void printArrayln(int[] arr , int si ,int ei)
{
    for(int i = si  ; i  <= ei ; i++)
    {
        out.print(arr[i] +" ") ;
    }
    out.println() ;
}


public static void printLArray(long[] arr , int si , int ei)
{
    for(int i = si ; i  <= ei ; i++)
    {
        out.print(arr[i] +" ") ;
    }
   
}




public static void printLArrayln(long[] arr , int si , int ei)
{
    for(int i = si ; i  <= ei ; i++)
    {
        out.print(arr[i] +" ") ;
    }
    out.println() ;
   
}

public static void printtwodArray(int[][] ans)
{
    for(int i = 0; i< ans.length ; i++)
    {
        for(int j  = 0 ; j <  ans[0].length ; j++)out.print(ans[i][j] +" ");
        out.println() ;
    }
    out.println() ;
   
}

  
   static long modPow(long a, long x, long p) {
    //calculates a^x mod p in logarithmic time.
    
    a = a % p ;
    
    if(a == 0)return 0L ;
    
    
    long res = 1L;
    while(x > 0) {
        if( x % 2 != 0) {
            res = (res * a) % p;
        }
        a = (a * a) % p;
        x =x/2;
    }
    return res;
}
 
 
 
  static long modInverse(long a, long p) {
    //calculates the modular multiplicative of a mod p.
    //(assuming p is prime).
    return modPow(a, p-2, p);
}
 
 static long[] factorial =  new long[1000001] ;
 
 static void modfac(long mod)
 {
     factorial[0]=1L ; factorial[1]=1L ;
     
     for(int i = 2; i<= 1000000 ;i++)
     {
         factorial[i] = factorial[i-1] *(long)(i) ;
          factorial[i] = factorial[i] % mod ;
     }
    
     
 }
 

 
 
 
static long modBinomial(long n, long r, long p) {
// calculates C(n,r) mod p (assuming p is prime).
 
  if(n < r) return 0L ; 
 
    long num = factorial[(int)(n)] ;
    
    long den = (factorial[(int)(r)]*factorial[(int)(n-r)]) %  p ;
    
    
    long ans  = num*(modInverse(den,p)) ;
    
    ans = ans % p ;
    
    return ans  ;
 
    
}
 
 
 static void update(int val , long[] bit ,int n)
 {
     for( ; val <= n ; val += (val &(-val)) )
     {
         bit[val]++ ;
     }
    
     
 }
 
 
 static long query(int val , long[] bit , int n)
 {
     long ans = 0L; 
     for( ; val >=1 ; val-=(val&(-val))  )ans += bit[val];
     
     return ans ;
 }



static int countSetBits(long n) 
    { 
        int count = 0; 
        while (n > 0) { 
            n = (n) & (n - 1L); 
            count++; 
        } 
        return count; 
    } 


static int abs(int x)
{
    if(x < 0)x = -1*x ;
    
    return x ;
}


static long abs(long x)
{
    if(x < 0)x = -1L*x ;
    
    return x ;

}

////////////////////////////////////////////////////////////////////////////////////////////////
static void p(int val)
{
    out.print(val) ;
}

static void p()
{
    out.print(" ") ;
}

static void pln(int val)
{
    out.println(val) ;
}

static void pln()
{
    out.println() ;
}


static void p(long val)
{
    out.print(val) ;
}



static void pln(long val)
{
    out.println(val) ;
}


static void yes()
{
    out.println("YES") ;
}
static void no()
{
    out.println("NO") ;
}

////////////////////////////////////////////////////////////////////////////////////////////

 // calculate total no of nos greater than or equal to key in  sorted array arr
 
static  int bs(int[] arr, int s ,int e  ,int key)
{
    if( s> e)return 0 ;
    
     int mid = (s+e)/2 ;
        
        if(arr[mid] <key)
        {
            return bs(arr ,mid+1,e , key) ;
            
        }
        
        
        else{
            
             return bs(arr ,s ,mid-1, key)  + e-mid+1;
            
            
        }
}
 
// static ArrayList<Integer>[] adj ;
// static int mod= 1000000007 ;


 



//////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////

public static void solve()
{
FastReader scn = new FastReader() ;

//Scanner scn = new Scanner(System.in);
//int[] store = {2 ,3, 5 , 7  ,11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 } ;

// product of first 11 prime nos is greater than 10 ^ 12;
//sieve() ;
//ArrayList<Integer> arr[] = new ArrayList[n] ;
ArrayList<Integer> list = new ArrayList<>() ;
ArrayList<Long> lista = new ArrayList<>() ;
ArrayList<Long> listb = new ArrayList<>() ;
// ArrayList<Integer> lista = new ArrayList<>() ;
// ArrayList<Integer> listb = new ArrayList<>() ;
//ArrayList<String> lists = new ArrayList<>() ;

HashMap<Integer,Integer> map = new HashMap<>() ;
//HashMap<Long,Long> map = new HashMap<>() ;
HashMap<Integer,Integer> mapx  = new HashMap<>() ;
HashMap<Integer,Integer> mapy  = new HashMap<>() ;
//HashMap<String,Integer> maps = new HashMap<>() ;
//HashMap<Integer,Boolean> mapb = new HashMap<>() ;
//HashMap<Point,Integer> point = new HashMap<>() ; 

 Set<Integer> set = new HashSet<>() ;
 Set<Integer> setx = new HashSet<>() ;
 Set<Integer> sety = new HashSet<>() ;

StringBuilder sb =new StringBuilder("") ;

//Collections.sort(list);

//if(map.containsKey(arr[i]))map.put(arr[i] , map.get(arr[i]) +1 ) ;
//else map.put(arr[i],1) ;

// if(map.containsKey(temp))map.put(temp , map.get(temp) +1 ) ;
// else map.put(temp,1) ;

//int bit  =Integer.bitCount(n);
// gives total no of set bits in n;

// Arrays.sort(arr, new Comparator<Pair>() {
// 			@Override
// 			public int compare(Pair a, Pair b) {
// 				if (a.first != b.first) {
// 					return a.first - b.first; // for increasing order of first
// 				}
// 				return a.second - b.second ; //if first is same then sort on second basis
// 			}
// 		});


int testcase = 1;
// testcase = scn.nextInt() ;
for(int testcases =1  ; testcases <= testcase ;testcases++)
{
    
 //if(map.containsKey(arr[i]))map.put(arr[i],map.get(arr[i])+1) ;else map.put(arr[i],1) ;
 //if(map.containsKey(temp))map.put(temp,map.get(temp)+1) ;else map.put(temp,1) ;
 
 //adj = new ArrayList[n] ;

// for(int i = 0; i< n; i++)
// {
//     adj[i] = new ArrayList<Integer>();
// }

 long n = scn.nextLong() ;long k = scn.nextLong() ;


if(k*(k-1)< n )out.println("NO");

else{
    
  out.println("YES");  
long count = 0l ;    
    
for(int i = 1; i <= k ;i++)
{
    for(int j =i+1; j <= k ; j++)
    {
        out.println(i+" " + j) ;
        count++ ;
        if(count == n)
        {
            out.flush() ;return  ;
        }
        out.println(j+" " + i) ;
        count++ ;
        
        if(count == n)
        {
            out.flush() ;return  ;
        }
    }
}
    
    
    
}




//out.println(ans) ;
//out.println(ans+" "+in) ;

//out.println("Case #" + testcases + ": "  + ans ) ;
//out.println("@") ;
set.clear() ;
sb.delete(0 , sb.length()) ;
list.clear() ;lista.clear() ;listb.clear() ;
map.clear() ;
mapx.clear() ;
mapy.clear() ;
setx.clear() ;sety.clear() ;

} // test case  end loop


out.flush() ;  
} // solve fn ends


public static void main (String[] args) throws java.lang.Exception
{
  

solve() ;
      
}


}
  
 class Pair 
{
  int first ;
  
int second  ;
  
  
 
    
      
@Override
	public String toString() {
	
	String ans = "" ;
	ans  += this.first ;
	ans += " ";
	ans += this.second ;
	
	return ans  ;
	}


}