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 N = 3e5 + 5, M = 3e5 + 5;
int a[M], b[M], c[N];
inline int rd() {
  int sl = 0, fh = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') sl = sl * 10 + ch - '0', ch = getchar();
  return sl * fh;
}
int main() {
  int n, m, x;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &a[i], &b[i]);
  }
  x = a[0];
  int num = 0;
  for (int i = 0; i < m; ++i) {
    if (a[i] == x || b[i] == x) {
      continue;
    }
    num++;
    c[a[i]]++;
    c[b[i]]++;
  }
  if (!num) {
    cout << "YES" << endl;
    return 0;
  } else {
    for (int i = 1; i <= n; ++i) {
      if (c[i] == num) {
        cout << "YES" << endl;
        return 0;
      }
    }
  }
  memset(c, 0, sizeof(c));
  x = b[0];
  num = 0;
  for (int i = 0; i < m; ++i) {
    if (a[i] == x || b[i] == x) {
      continue;
    }
    num++;
    c[a[i]]++;
    c[b[i]]++;
  }
  if (!num) {
    cout << "YES" << endl;
    return 0;
  } else {
    for (int i = 1; i <= n; ++i) {
      if (c[i] == num) {
        cout << "YES" << endl;
        return 0;
      }
    }
    cout << "NO" << endl;
    return 0;
  }
}
