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 + 10;
int n, m, a, b, cnt[maxn];
vector<pair<int, int> > v;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    v.push_back(make_pair(a, b));
  }
  int now = v[0].first;
  int flag = 0;
  memset(cnt, 0, sizeof(cnt));
  cnt[now] = 1;
  for (int i = 1; i < m; i++) {
    if (v[i].first == now || v[i].second == now) {
      cnt[now]++;
    } else {
      cnt[v[i].first]++;
      cnt[v[i].second]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i != now) {
      if (cnt[i] + cnt[now] == m) flag = 1;
    }
  }
  now = v[0].second;
  memset(cnt, 0, sizeof(cnt));
  cnt[now] = 1;
  for (int i = 1; i < m; i++) {
    if (v[i].first == now || v[i].second == now) {
      cnt[now]++;
    } else {
      cnt[v[i].first]++;
      cnt[v[i].second]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i != now) {
      if (cnt[i] + cnt[now] == m) flag = 1;
    }
  }
  if (flag == 1)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}
