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(0);
  long long a, b, c = 0, d, e = 0, f, i, j, k, m, n, o, x, y;
  cin >> n >> m;
  map<pair<int, int>, int> ok;
  long long cnt[n + 1];
  memset(cnt, 0, sizeof(cnt));
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    ok[{a, b}]++;
    ok[{b, a}]++;
    cnt[a]++;
    cnt[b]++;
  }
  for (i = n; i > 0; i--) {
    if (cnt[i] > cnt[e]) e = i;
  }
  for (i = n; i > 0; i--) {
    if (i == e) continue;
    if (cnt[e] + cnt[i] - ok[{e, i}] >= m) c++;
  }
  f = e;
  e = 0;
  if (c == 0) {
    for (i = n; i > 0; i--) {
      if (i == f) continue;
      if (cnt[i] > cnt[e]) e = i;
    }
    for (i = n; i > 0; i--) {
      if (i == e) continue;
      if (cnt[e] + cnt[i] - ok[{e, i}] >= m) c++;
    }
  }
  if (c > 0)
    cout << "YES\n";
  else
    cout << "NO\n";
}
