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 = 1e6 + 10;
int n, m, l[maxn], r[maxn];
bool ok, f[maxn];
bool ck(int x) {
  memset(f, 0, sizeof f);
  int now1 = 0, now2 = 0, a, b;
  for (int i = 2; i <= m; i++)
    if (l[i] == x || r[i] == x) f[i] = 1, now1++, now2++;
  for (int i = 2; i <= m; i++)
    if (!f[i]) {
      a = l[i];
      b = r[i];
      break;
    }
  for (int i = 2; i <= m; i++)
    if (!f[i]) {
      if (l[i] == a || r[i] == a) now1++;
      if (l[i] == b || r[i] == b) now2++;
    }
  return (now1 == m - 1) || (now2 == m - 1);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d%d", &l[i], &r[i]);
  ok = (ok || ck(l[1]));
  ok = (ok || ck(r[1]));
  printf(ok ? "YES\n" : "NO\n");
  return 0;
}
