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.io.*;

public class tr0 {
	static PrintWriter out;
	static StringBuilder sb;
	static final double EPS = 1e-9;
	static long mod = 1000000007;
	static int inf = (int) 1e9 + 2;
	static long[] fac;
	static Long[] a;
	static int[] si;
	static ArrayList<Integer> primes;
	static ArrayList<Integer>[] ad;
	static ArrayList<pair>[] d;
	static edge[] ed;
	static int[][] vis;
	static int[] l, ch;
	static int[] occ, p;
	static Queue<Integer>[] can;
	static String s;
	static long[][][] memo;
	static int n, m, k;
	static qu[] ar;
	static long[] sz;
	static long ans = 0;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		long n = sc.nextInt();
		long k = sc.nextInt();
		if (k * (k - 1) < n) {
			System.out.println("NO");
			return;
		}
		Stack<pair> s = new Stack<>();
		int now =(int) k;
		int l = now - 1;
		boolean tr = true;
		while (n > 0) {
			if (l == 0) {
				now--;
				l = now - 1;
				if (!s.isEmpty()) {
                     if((int)s.peek().to==now && tr)
                    	 tr=!tr;
                     else if(tr && (int)s.peek().number==l)
                    	 tr=!tr;
                     else if(!tr && (int)s.peek().to==l)
                    	 tr=!tr;
                     else if(!tr && (int)s.peek().number==now)
                    	 tr=!tr;
				}
			}
			if(l==0)
				break;
			if (tr) {
				s.add(new pair(now, now - l));
				if(n==1)
					break;
				s.add(new pair(now - l, now));
			} else {
				s.add(new pair(now - l, now));
				if(n==1)
					break;
				s.add(new pair(now, now - l));
			}
                n-=2;
			l--;
		}
		out.println("YES");
		while (!s.isEmpty())
			out.println(s.pop());
		out.flush();
	}

	static class qu implements Comparable<qu> {
		long a, b, c;

		qu(long i, long v, long x) {
			a = i;
			b = v;
			c = x;
		}

		public String toString() {
			return a + " " + b + " " + c;
		}

		public int compareTo(qu o) {
			if (Math.abs(o.a - o.b) <= Math.abs(a - b) && Math.abs(o.a - o.b) <= Math.abs(a - c))
				return 1;
			if (Math.abs(o.a - o.c) <= Math.abs(a - b) && Math.abs(o.a - o.c) <= Math.abs(a - c))
				return 1;
			else
				return -1;
		}
	}

	static class seg implements Comparable<seg> {
		int a;
		int b;
		int l;
		int r;
		int bit;

		seg(int s, int e, int x, int y, int bi) {
			a = s;
			b = e;
			l = x;
			r = y;
			bit = bi;
		}

		public String toString() {
			return a + " " + b + " " + l + " " + r + " " + bit;
		}

		public int compareTo(seg o) {
			// if(a==o.a)
			return bit - o.bit;
			// return
		}
	}

	static class pair {
		long to;
		long number;

		pair(long t, long n) {
			number = n;
			to = t;
		}

		public String toString() {
			return to + " " + number;
		}
	}

	static long modPow(long a, int e)

	{
		long res = 1;
		while (e > 0) {
			if ((e & 1) == 1)
				res = (res * a) % mod;
			a = (a * a) % mod;
			e >>= 1;
		}
		return res;
	}

	static long inver(long x) {
		long a = x;
		long e = (mod - 2);
		long res = 1;
		while (e > 0) {
			if ((e & 1) == 1) {
				res = ((1l * res * a) % mod);
			}
			a = ((1l * a * a) % mod);
			e >>= 1;
		}
		return res % mod;
	}

	static class edge implements Comparable<edge> {
		int from;
		int to;
		int number;

		edge(int f, int t, int n) {
			from = f;
			to = t;
			number = n;
		}

		public String toString() {
			return from + " " + to + " " + number;
		}

		public int compareTo(edge f) {
			return f.number - number;
		}
	}

	static void seive(int N) {
		si = new int[N];
		primes = new ArrayList<>();
		si[1] = 1;
		for (int i = 2; i < N; i++) {
			if (si[i] == 0) {
				si[i] = i;
				primes.add(i);
			}
			for (int j = 0; j < primes.size() && primes.get(j) <= si[i] && (i * primes.get(j)) < N; j++)
				si[primes.get(j) * i] = primes.get(j);

		}
	}

	static long fac(int n) {
		if (n == 0)
			return fac[n] = 1;
		if (n == 1)
			return fac[n] = 1;
		long ans = 1;
		for (int i = 1; i <= n; i++)
			fac[i] = ans = (i % mod * ans % mod) % mod;
		return ans % mod;
	}

	static long gcd(long a, long b) {

		if (b == 0)
			return a;
		return gcd(b, a % b);
	}

	static class unionfind {
		int[] p;
		int[] size;
		int[] max;
		int num;

		unionfind(int n) {
			p = new int[n];
			size = new int[n];
			max = new int[n];
			for (int i = 0; i < n; i++) {
				p[i] = i;
				max[i] = i;
			}
			Arrays.fill(size, 1);
			num = n;
		}

		int findSet(int v) {
			if (v == p[v])
				return v;

			max[v] = Math.max(max[v], max[p[v]]);

			p[v] = findSet(p[v]);

			max[v] = Math.max(max[v], max[p[v]]);

			return p[v];
		}

		boolean sameSet(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return true;
			return false;
		}

		int max() {
			int max = 0;
			for (int i = 0; i < size.length; i++)
				if (size[i] > max)
					max = size[i];
			return max;
		}

		boolean combine(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return true;
			// System.out.println(num+" ppp");
			num--;
			if (size[a] > size[b]) {
				p[b] = a;
				max[a] = Math.max(max[a], max[b]);
				size[a] += size[b];

			} else {
				p[a] = b;
				max[b] = Math.max(max[a], max[b]);
				size[b] += size[a];
			}
			return false;
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}