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 init = 1e9 + 7;
const long long N = 3e5 + 7;
vector<int> counts1(N, 0);
vector<pair<int, int> > v;
int m;
string solve(int x) {
  int c = 1, maxc = -1;
  for (int i = 1; i < m; ++i) {
    if (v[i].first == x || v[i].second == x) {
      ++c;
    } else {
      ++counts1[v[i].first];
      ++counts1[v[i].second];
      maxc = max(maxc, max(counts1[v[i].first], counts1[v[i].second]));
    }
  }
  if (c == m || maxc == m - c) {
    return "YES";
  }
  return "NO";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    v.push_back(make_pair(a, b));
  }
  string ans = solve(v[0].first);
  if (ans == "YES")
    cout << ans << endl;
  else {
    for (int i = 0; i <= n; ++i) counts1[i] = 0;
    ans = solve(v[0].second);
    cout << ans << endl;
  }
  return 0;
}
