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>
const int MAXN = (int)3e5 + 5;
std::pair<int, int> p[MAXN];
int main() {
  int n, m;
  std::cin >> n >> m;
  for (int i = 1; i <= m; i++) scanf("%d%d", &p[i].first, &p[i].second);
  int x, y, copyi;
  int i;
  bool flag = false;
  x = p[1].first;
  for (i = 1; i <= m; i++) {
    if (x != p[i].second && x != p[i].first) break;
  }
  if (i > m)
    flag = true;
  else {
    copyi = i;
    y = p[i].first;
    for (; copyi <= m; copyi++) {
      if (y != p[copyi].second && y != p[copyi].first && x != p[copyi].second &&
          x != p[copyi].first)
        break;
    }
    if (copyi > m) flag = true;
    copyi = i;
    y = p[i].second;
    for (; copyi <= m; copyi++) {
      if (y != p[copyi].second && y != p[copyi].first && x != p[copyi].second &&
          x != p[copyi].first)
        break;
    }
    if (copyi > m) flag = true;
  }
  x = p[1].second;
  for (i = 1; i <= m; i++) {
    if (x != p[i].second && x != p[i].first) break;
  }
  if (i > m)
    flag = true;
  else {
    copyi = i;
    y = p[i].first;
    for (; copyi <= m; copyi++) {
      if (y != p[copyi].second && y != p[copyi].first && x != p[copyi].second &&
          x != p[copyi].first)
        break;
    }
    if (copyi > m) flag = true;
    copyi = i;
    y = p[i].second;
    for (; copyi <= m; copyi++) {
      if (y != p[copyi].second && y != p[copyi].first && x != p[copyi].second &&
          x != p[copyi].first)
        break;
    }
    if (copyi > m) flag = true;
  }
  if (flag)
    printf("YES");
  else
    printf("NO");
}
