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 = 3e5 + 10;
int n, m;
struct Node {
  int a;
  int b;
};
Node node[N];
bool cmp(Node A, Node B) {
  if (A.a == B.a) return A.b < B.b;
  return A.a < B.a;
}
int check(int x, int y) {
  for (int i = 0; i < m; i++) {
    if (x != node[i].a && x != node[i].b && y != node[i].a && y != node[i].b)
      return 0;
  }
  return 1;
}
int main() {
  int r1, r2, r3, r4;
  while (scanf("%d%d", &n, &m) != EOF) {
    r1 = 0, r2 = 0, r3 = 0, r4 = 0;
    for (int i = 0; i < m; i++) {
      scanf("%d%d", &node[i].a, &node[i].b);
      if (i == 0) {
        r1 = node[0].a;
        r2 = node[0].b;
      } else {
        if (node[i].a != r1 && node[i].b != r1 && node[i].a != r2 &&
            node[i].b != r2) {
          r3 = node[i].a;
          r4 = node[i].b;
        }
      }
    }
    int one = (check(r1, r2) || check(r1, r3) || check(r1, r4) ||
               check(r2, r3) || check(r2, r4));
    if (one)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
