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;
struct node {
  int a, b;
} nn[300005];
bool check(int a, int b) {
  for (int i = 0; i < m; i++) {
    if (nn[i].a != a && nn[i].b != a && nn[i].a != b && nn[i].b != b)
      return false;
  }
  return true;
}
int main() {
  scanf("%d %d", &n, &m);
  int a, b, c, d;
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &nn[i].a, &nn[i].b);
    if (i == 0) a = nn[i].a, b = nn[i].b;
    if (nn[i].a != a && nn[i].b != a && nn[i].a != b && nn[i].b != b)
      c = nn[i].a, d = nn[i].b;
  }
  int f = 0;
  if (check(a, b) || check(a, c) || check(a, d) || check(b, c) || check(b, d))
    f = 1;
  if (f)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
