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;
void solve() {
  long long int n, m, i, j;
  cin >> n >> m;
  vector<pair<long long int, long long int>> vec;
  for (i = 0; i < m; i++) {
    long long int a, b;
    cin >> a >> b;
    vec.push_back({a, b});
  }
  long long int a = vec[0].first;
  long long int f = 0;
  for (i = 1; i < m; i++) {
    long long int x, y;
    x = vec[i].first;
    y = vec[i].second;
    if (x != a && y != a) {
      long long int b = vec[i].first;
      for (j = i + 1; j < m; j++) {
        long long int xt, yt;
        xt = vec[j].first;
        yt = vec[j].second;
        if (xt != a && xt != b && yt != a && yt != b) {
          break;
        }
      }
      if (j >= m) {
        cout << "YES\n";
        return;
      }
      b = vec[i].second;
      for (j = i + 1; j < m; j++) {
        long long int xt, yt;
        xt = vec[j].first;
        yt = vec[j].second;
        if (xt != a && xt != b && yt != a && yt != b) {
          break;
        }
      }
      if (j >= m) {
        cout << "YES\n";
        return;
      }
      f = 1;
    }
    if (f) break;
  }
  if (i >= m) {
    cout << "YES\n";
    return;
  }
  a = vec[0].second;
  f = 0;
  for (i = 1; i < m; i++) {
    long long int x, y;
    x = vec[i].first;
    y = vec[i].second;
    if (x != a && y != a) {
      long long int b = vec[i].first;
      for (j = i + 1; j < m; j++) {
        long long int xt, yt;
        xt = vec[j].first;
        yt = vec[j].second;
        if (xt != a && xt != b && yt != a && yt != b) {
          break;
        }
      }
      if (j >= m) {
        cout << "YES\n";
        return;
      }
      b = vec[i].second;
      for (j = i + 1; j < m; j++) {
        long long int xt, yt;
        xt = vec[j].first;
        yt = vec[j].second;
        if (xt != a && xt != b && yt != a && yt != b) {
          break;
        }
      }
      if (j >= m) {
        cout << "YES\n";
        return;
      }
      f = 1;
    }
    if (f) break;
  }
  if (i >= m) {
    cout << "YES\n";
    return;
  }
  cout << "NO\n";
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(NULL);
  int t;
  t = 1;
  while (t > 0) {
    t--;
    solve();
  }
}
