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>
struct node {
  int x, y;
} a[300005];
int n, m;
bool judge(int x, int y) {
  int flag = 0;
  for (int i = 0; i < m; i++) {
    if (a[i].x != x && a[i].y != y && a[i].x != y && a[i].y != x) {
      return false;
    }
  }
  return true;
}
int main() {
  int x1, x2, y1, y2;
  int i;
  scanf("%d%d", &n, &m);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
  }
  x1 = a[0].x;
  y1 = a[0].y;
  int flag = 1;
  for (i = 1; i < m; i++) {
    if (a[i].x != x1 && a[i].y != x1 && a[i].x != y1 && a[i].y != y1) {
      x2 = a[i].x;
      y2 = a[i].y;
      flag = 0;
      break;
    }
  }
  if (flag == 1) {
    printf("YES\n");
  } else {
    if (judge(x1, x2) || judge(y1, y2) || judge(x1, y2) || judge(x2, y1) ||
        judge(x1, y1) || judge(x2, y2))
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
