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>
using namespace std;
int n, m, a[300005][2];
inline bool judge(int x, int y) {
  for (int i = 1; i <= m; i++) {
    if (a[i][0] != x && a[i][1] != x && a[i][0] != y && a[i][1] != y) {
      return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &a[1][0], &a[1][1]);
  int k = 1;
  for (int i = 2; i <= m; i++) {
    scanf("%d%d", &a[i][0], &a[i][1]);
    if (a[1][0] == a[k][0] || a[1][0] == a[k][1] || a[1][1] == a[k][0] ||
        a[1][1] == a[k][1]) {
      k = i;
    }
  }
  if (judge(a[1][0], a[k][0]) || judge(a[1][1], a[k][0]) ||
      judge(a[1][0], a[k][1]) || judge(a[1][1], a[k][1])) {
    printf("YES");
  } else {
    printf("NO");
  }
  return 0;
}
