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 maxn = 3e5 + 5;
int n, m;
int first[maxn], second[maxn];
void solve() {
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) cin >> first[i] >> second[i];
  int a, b;
  a = first[1];
  b = -1;
  for (int i = 1; i <= m; ++i) {
    if (first[i] != a && second[i] != a) {
      if (first[i] != b && second[i] != b) {
        if (b == -1)
          b = first[i];
        else
          break;
      }
    }
    if (i == m) {
      puts("YES");
      return;
    }
  }
  a = -1;
  b = second[1];
  for (int i = 1; i <= m; ++i) {
    if (first[i] != b && second[i] != b) {
      if (first[i] != a && second[i] != a) {
        if (a == -1)
          a = first[i];
        else
          break;
      }
    }
    if (i == m) {
      puts("YES");
      return;
    }
  }
  a = -1;
  b = second[1];
  for (int i = 1; i <= m; ++i) {
    if (first[i] != b && second[i] != b) {
      if (first[i] != a && second[i] != a) {
        if (a == -1)
          a = second[i];
        else
          break;
      }
    }
    if (i == m) {
      puts("YES");
      return;
    }
  }
  a = first[1];
  b = -1;
  for (int i = 1; i <= m; ++i) {
    if (first[i] != a && second[i] != a) {
      if (first[i] != b && second[i] != b) {
        if (b == -1)
          b = second[i];
        else
          break;
      }
    }
    if (i == m) {
      puts("YES");
      return;
    }
  }
  puts("NO");
}
int main() { solve(); }
