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 + 9;
int a[maxn][2], vis[maxn], mp[maxn];
vector<int> vec;
bool cmp(int a, int b) { return a > b; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie();
  int i, j, k, n, m;
  cin >> n >> m;
  for (i = 0; i < m; i++) {
    for (j = 0; j < 2; j++) {
      cin >> a[i][j];
    }
  }
  int x = a[0][0], cnt = 1;
  for (i = 1; i < m; i++) {
    int flag = 0;
    for (j = 0; j < 2; j++) {
      if (a[i][j] == x) flag = 1;
    }
    if (flag) vis[i] = 1, cnt++;
  }
  for (i = 1; i < m; i++) {
    if (!vis[i]) {
      for (j = 0; j < 2; j++) {
        mp[a[i][j]]++;
      }
    }
  }
  for (j = 0; j <= n; j++) {
    vec.push_back(mp[j]);
  }
  sort(vec.begin(), vec.end(), cmp);
  if (vec[0] >= m - cnt) {
    cout << "YES" << endl;
    return 0;
  }
  vec.clear();
  memset(vis, 0, sizeof(vis));
  memset(mp, 0, sizeof(mp));
  x = a[0][1], cnt = 1;
  for (i = 1; i < m; i++) {
    int flag = 0;
    for (j = 0; j < 2; j++) {
      if (a[i][j] == x) flag = 1;
    }
    if (flag) vis[i] = 1, cnt++;
  }
  for (i = 1; i < m; i++) {
    if (!vis[i]) {
      for (j = 0; j < 2; j++) {
        mp[a[i][j]]++;
      }
    }
  }
  for (j = 0; j <= n; j++) {
    vec.push_back(mp[j]);
  }
  sort(vec.begin(), vec.end(), cmp);
  if (vec[0] >= m - cnt) {
    cout << "YES" << endl;
    return 0;
  }
  cout << "NO" << endl;
}
