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 main() {
  int n;
  int m;
  cin >> n >> m;
  pair<int, int> a[m];
  for (int i = 0; i < m; ++i) {
    cin >> a[i].first >> a[i].second;
  }
  int x = a[0].first, x1 = a[0].second;
  vector<pair<int, int> > b[2];
  for (int i = 1; i < m; ++i) {
    if (a[i].first != x && a[i].second != x) b[0].push_back(a[i]);
  }
  for (int i = 1; i < m; ++i) {
    if (a[i].first != x1 && a[i].second != x1) b[1].push_back(a[i]);
  }
  if (b[0].size() == 0 || b[1].size() == 0) {
    cout << "YES";
    exit(0);
  }
  int nans = 0;
  for (auto g : b) {
    int y1 = g[0].first, y2 = g[0].second;
    for (int i = 1; i < g.size(); ++i) {
      if (y1 == g[i].first) {
        if (y2 != g[i].second) y2 = -1;
      } else if (y1 == g[i].second) {
        if (y2 != g[i].first) {
          y2 = -1;
        }
      } else {
        y1 = -1;
        if (y2 != g[i].second && y2 != g[i].first) {
          nans++;
          break;
        }
      }
    }
  }
  if (nans < 2)
    cout << "YES";
  else
    cout << "NO";
}
