Toad Ivan has m pairs of integers, each integer is between 1 and n, inclusive. The pairs are (a_1, b_1), (a_2, b_2), …, (a_m, b_m). 

He asks you to check if there exist two integers x and y (1 ≤ x < y ≤ n) such that in each given pair at least one integer is equal to x or y.

Input

The first line contains two space-separated integers n and m (2 ≤ n ≤ 300 000, 1 ≤ m ≤ 300 000) — the upper bound on the values of integers in the pairs, and the number of given pairs.

The next m lines contain two integers each, the i-th of them contains two space-separated integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the integers in the i-th pair.

Output

Output "YES" if there exist two integers x and y (1 ≤ x < y ≤ n) such that in each given pair at least one integer is equal to x or y. Otherwise, print "NO". You can print each letter in any case (upper or lower).

Examples

Input


4 6
1 2
1 3
1 4
2 3
2 4
3 4


Output


NO


Input


5 4
1 2
2 3
3 4
4 5


Output


YES


Input


300000 5
1 2
1 2
1 2
1 2
1 2


Output


YES

Note

In the first example, you can't choose any x, y because for each such pair you can find a given pair where both numbers are different from chosen integers.

In the second example, you can choose x=2 and y=4.

In the third example, you can choose x=1 and y=2.
#include <bits/stdc++.h>
int main(void) {
  long int n, m;
  scanf(" %ld %ld", &n, &m);
  long int a[m], b[m];
  for (long int i = 0; i < m; i++) {
    scanf(" %ld %ld", &a[i], &b[i]);
  }
  int temp = a[0];
  int flag = 0;
  for (long int i = 1; i < m; i++) {
    if (a[i] != temp && b[i] != temp) {
      int flag1 = 0;
      for (long int j = 0; j < m; j++) {
        if (a[j] != temp && b[j] != temp && a[j] != a[i] && b[j] != a[i]) {
          flag1 = 1;
          break;
        }
      }
      if (flag1 == 0) {
        printf("YES");
        return 0;
      }
      flag1 = 0;
      for (long int j = 0; j < m; j++) {
        if (a[j] != temp && b[j] != temp && a[j] != b[i] && b[j] != b[i]) {
          flag1 = 1;
          break;
        }
      }
      if (flag1 == 0) {
        printf("YES");
        return 0;
      }
      flag = 1;
      break;
    }
  }
  if (flag == 0) {
    printf("YES");
    return 0;
  }
  temp = b[0];
  flag = 0;
  for (long int i = 1; i < m; i++) {
    if (a[i] != temp && b[i] != temp) {
      int flag1 = 0;
      for (long int j = 0; j < m; j++) {
        if (a[j] != temp && b[j] != temp && a[j] != a[i] && b[j] != a[i]) {
          flag1 = 1;
          break;
        }
      }
      if (flag1 == 0) {
        printf("YES");
        return 0;
      }
      flag1 = 0;
      for (long int j = 0; j < m; j++) {
        if (a[j] != temp && b[j] != temp && a[j] != b[i] && b[j] != b[i]) {
          flag1 = 1;
          break;
        }
      }
      if (flag1 == 0) {
        printf("YES");
        return 0;
      }
      flag = 1;
      break;
    }
  }
  if (flag == 0) {
    printf("YES");
  } else {
    printf("NO");
  }
  return 0;
}
