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;
const int N = 300005;
int n, m, a[N][3], dif, ans, x, y;
int valid(int x, int i) { return a[i][1] == x || a[i][2] == x; }
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) scanf("%d%d", &a[i][1], &a[i][2]);
  for (int j = 1; j <= 2; ++j) {
    dif = 0;
    ans = 1;
    for (int i = 2; i <= m; ++i)
      if (!valid(a[1][j], i)) {
        if (dif == 0) {
          dif = 1;
          x = a[i][1];
          y = a[i][2];
        } else {
          if (!valid(x, i)) x = y;
          if (!valid(y, i)) y = x;
          if (!valid(x, i) && !valid(y, i)) {
            ans = 0;
            break;
          }
        }
      }
    if (ans) {
      puts("YES");
      return 0;
    }
  }
  puts("NO");
}
