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.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer; 
import java.util.*;
import java.io.*;
public class codeforces 
{ 
   static class Student{
       int x,y;
        Student(int x,int y){
            this.x=x;
            this.y=y;
            //this.z=z;
        }
    }
    static int prime[];
    static void sieveOfEratosthenes(int n) 
    { 
        // Create a boolean array "prime[0..n]" and initialize 
        // all entries it as true. A value in prime[i] will 
        // finally be false if i is Not a prime, else true. 
        int pos=0;
        prime= new int[n+1]; 
        for(int p = 2; p*p <=n; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == 0)
            { 
                // Update all multiples of p
                prime[p]=p;
                for(int i = p*p; i <= n; i += p)
                    if(prime[i]==0)
                    prime[i] = p; 
            } 
        } 
    }
static    class Sortbyroll implements Comparator<Student> 
{ 
    // Used for sorting in ascending order of 
    // roll number 
    public int compare(Student c, Student b) 
    { 
        return c.y-b.y;
    } 
} 
static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        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; 
        } 
    } 
 
    static class Edge{
        int a,b;
        Edge(int a,int b){
            this.a=a;
            this.b=b;
        }
    }
    static class Trie{
        HashMap<Character,Trie>map;
         int c;
        Trie(){
            map=new HashMap<>();
            //z=null;
            //o=null;
            c=0;
        }
    }
    //static long ans;
    static int parent[];
    static int rank[];
   // static int b[][];
    static int bo[];
    static int ho[];
    static int seg[];
    //static int pos;
     static long mod=1000000007;
    //static int dp[][];
    static HashMap<Integer,Integer>map;
    static PriorityQueue<Student>q=new PriorityQueue<>();
    //static Stack<Integer>st;
   // static ArrayList<Character>ans;
    static ArrayList<ArrayList<Integer>>adj;
    //static long ans;
    static int pos;
    static Trie root;
    static long fac[];
    static int gw,gb;
    //static long mod=(long)(998244353);
    //static int ans;
    static void solve()throws IOException{
        FastReader sc=new FastReader();
        int n,k,x,t,y,i;
        n=sc.nextInt();
        k=sc.nextInt();
        if((long)(k)*(long)(k-1)<n)
            System.out.println("NO");
        else{
            x=1;
            y=1;
            t=0;
            StringBuilder sb=new StringBuilder();
            while(t<n){
               ++x;
                y=1;
                for(i=1;i<=k;i++){
                    if(x>k)
                        x=1;
                    sb.append(y+" "+x+"\n");
                    ++y;
                    ++x;
                    ++t;
                    if(t>=n)
                        break;
                }
            }
            System.out.println("YES");
            System.out.println(sb.toString());
        }
    }
    static long fac(long x,long y){
        if(y==0||x==0)
            return 1;
        return x * fac(x-1,y-1);
    }
    static long nCr(long n, long r, 
                                    long p) 
    { 
        return (fac[(int)n]* modInverse(fac[(int)r], p) 
                % p * modInverse(fac[(int)(n-r)], p) 
                                    % p) % p; 
    } 
    //static int prime[];
    //static int dp[];
    public static void main(String[] args){
        //long sum=0;
        try {
                codeforces.solve();
                } catch (Exception e) {
                    e.printStackTrace();
                }
    }
       static long modInverse(long n, long p) 
    { 
        return power(n, p-(long)2,p); 
    } 
 static long power(long x, long y, long p) 
{ 
    long res = 1;      // Initialize result 
  
    x = x % p;  // Update x if it is more than or  
                // equal to p 
  
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if (y %(long)2!=0) 
            res = (res*x) % p; 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x) % p;   
    } 
    return res%p;
 }
    /*public static long power(long x,long a) {
		if(a==0) return 1;
		if(a%2==1)return (x*1L*power(x,a-1))%mod;
		return (power((int)((x*1L*x)%mod),a/2))%mod;
	}*/
   static int find(int x) 
    { 
        // Finds the representative of the set 
        // that x is an element of 
        while(parent[x]!=x) 
        { 
            // if x is not the parent of itself 
            // Then x is not the representative of 
            // his set,
            x=parent[x];
  
            // so we recursively call Find on its parent 
            // and move i's node directly under the 
            // representative of this set 
        } 
  
        return x; 
    }
    static void union(int x, int y) 
    { 
        // Find representatives of two sets 
        int xRoot = find(x), yRoot = find(y); 
  
        // Elements are in the same set, no need 
        // to unite anything. 
        if (xRoot == yRoot) 
            return; 
  
         // If x's rank is less than y's rank 
        if (rank[xRoot] < rank[yRoot]) 
  
            // Then move x under y  so that depth 
            // of tree remains less 
            parent[xRoot] = yRoot; 
  
        // Else if y's rank is less than x's rank 
        else if (rank[yRoot] < rank[xRoot]) 
  
            // Then move y under x so that depth of 
            // tree remains less 
            parent[yRoot] = xRoot; 
  
        else // if ranks are the same 
        { 
            // Then move y under x (doesn't matter 
            // which one goes where) 
            parent[yRoot] = xRoot; 
  
            // And increment the the result tree's 
            // rank by 1 
            rank[xRoot] = rank[xRoot] + 1; 
        } 
    } 
    static long gcd(long a, long b) 
    { 
        if (a == 0){ 
             //ans+=b;
            return b;
        }
        return gcd(b % a, a); 
    }
}