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 long long INF = 922337203685477110;
const long long N = 300005;
const long long mod = 1e9 + 7;
const long long M = 22;
vector<pair<long long, long long> > v;
long long vis[N];
bool check(long long y) {
  for (long long i = 0; i < v.size(); i++) {
    if (!vis[i]) continue;
    if (v[i].first != y && v[i].second != y) return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  while (m--) {
    long long a, b;
    cin >> a >> b;
    v.push_back({a, b});
  }
  long long x = v[0].first;
  long long y1 = -1, y2 = -1;
  for (long long i = 0; i < v.size(); i++) {
    if (v[i].first != x && v[i].second != x) {
      y1 = v[i].first;
      y2 = v[i].second;
      vis[i] = 1;
    }
  }
  if (check(y1) || check(y2))
    cout << "YES"
         << "\n";
  else {
    x = v[0].second;
    y1 = -1, y2 = -1;
    memset(vis, 0, sizeof(vis));
    for (long long i = 0; i < v.size(); i++) {
      if (v[i].first != x && v[i].second != x) {
        y1 = v[i].first;
        y2 = v[i].second;
        vis[i] = 1;
      }
    }
    if (check(y1) || check(y2))
      cout << "YES"
           << "\n";
    else
      cout << "NO"
           << "\n";
  }
}
