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;
map<int, int> cnt;
map<pair<int, int>, int> s;
bool flag;
int main() {
  int n, m, a, b;
  cin >> n >> m >> a >> b;
  if (m == 1) flag = true;
  s[make_pair(a, b)]++;
  s[make_pair(b, a)]++;
  cnt[a]++;
  cnt[b]++;
  for (int i = 2; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    s[make_pair(x, y)]++;
    s[make_pair(y, x)]++;
    cnt[x]++;
    cnt[y]++;
  }
  for (int i = 1; i <= n; i++) {
    int u = i;
    int v = a;
    if (u != v) {
      if (cnt[u] + cnt[v] - s[make_pair(u, v)] == m) flag = true;
    }
    v = b;
    if (u != v) {
      if (cnt[u] + cnt[v] - s[make_pair(u, v)] == m) flag = true;
    }
  }
  if (flag)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
