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() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n, m, a, b, var1, var2;
    cin >> n >> m;
    long long int i;
    vector<pair<long long int, long long int> > v;
    for (i = 0; i < m; i++) {
      cin >> a >> b;
      v.push_back(pair<long long int, long long int>(a, b));
    }
    for (i = 1; i < m; i++) {
      if (v[0].first != v[i].first && v[0].first != v[i].second) {
        var1 = v[i].first;
        var2 = v[i].second;
        break;
      }
    }
    long long int count1 = 0, count2 = 0, count = 1;
    for (i = 1; i < m; i++) {
      if (v[i].first != v[0].first && v[0].first != v[i].second) {
        if (v[i].first == var1 || v[i].second == var1) {
          count1++;
        }
        if (v[i].first == var2 || v[i].second == var2) {
          count2++;
          continue;
        }
        if (v[i].first != var1 && v[i].second != var1 && v[i].first != var2 &&
            v[i].second != var2) {
          goto sos;
        }
      } else {
        count++;
      }
    }
    if (count1 + count == m || count2 + count == m) {
      cout << "YES";
      return 0;
    }
  sos:
    for (i = 1; i < m; i++) {
      if (v[0].second != v[i].first && v[0].second != v[i].second) {
        var1 = v[i].first;
        var2 = v[i].second;
        break;
      }
    }
    count1 = 0, count2 = 0, count = 1;
    for (i = 1; i < m; i++) {
      if (v[i].first != v[0].second && v[0].second != v[i].second) {
        if (v[i].first == var1 || v[i].second == var1) {
          count1++;
        }
        if (v[i].first == var2 || v[i].second == var2) {
          count2++;
          continue;
        }
        if (v[i].first != var1 && v[i].second != var1 && v[i].first != var2 &&
            v[i].second != var2) {
          cout << "NO";
          return 0;
        }
      } else {
        count++;
      }
    }
    if (count1 + count == m || count2 + count == m) {
      cout << "YES";
      return 0;
    } else {
      cout << "NO";
      return 0;
    }
  }
}
