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 + 1;
int ai[N][2];
double ad[N];
long long all[N];
long double ald[N];
set<int> s;
int n, m, l, r;
int main() {
  cin >> n >> m;
  if (n < 3) {
    cout << "YES";
    return 0;
  }
  for (int i = 0; i < m; ++i) {
    cin >> ai[i][0] >> ai[i][1];
  }
  for (int j = 0; j < 2; ++j) {
    int x = ai[0][j];
    bool bi[2];
    bi[0] = true;
    bi[1] = true;
    int r = -1;
    for (int i = 1; i < m; ++i) {
      if (ai[i][0] == x || ai[i][1] == x) continue;
      if (r < 0) r = i;
      for (int k = 0; k < 2; ++k) {
        if (ai[r][k] != ai[i][0] && ai[r][k] != ai[i][1]) bi[k] = false;
      }
    }
    if (bi[0] || bi[1]) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
  return 0;
}
