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 + 1000;
vector<int> mp[N];
set<int> ss[N];
int x[N], y[N];
int a, b;
int n, m;
int check(int i, int q, int p) {
  int f = 1;
  for (; i <= m; i++) {
    if (x[i] == q || y[i] == q) continue;
    if (x[i] == p || y[i] == p) continue;
    f = 0;
  }
  return f;
}
int main() {
  scanf("%d%d", &n, &m);
  int flag = 0;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x[i], &y[i]);
  }
  if (m <= 2) {
    puts("YES");
    return 0;
  }
  a = x[1];
  for (int i = 2; i <= m; i++) {
    if (x[i] == a || y[i] == a) {
      if (i == m) flag = 1;
      continue;
    }
    if (check(i + 1, a, x[i])) flag = 1;
    if (check(i + 1, a, y[i])) flag = 1;
    break;
  }
  if (flag) {
    puts("YES");
    return 0;
  }
  a = y[1];
  for (int i = 2; i <= m; i++) {
    if (x[i] == a || y[i] == a) {
      if (i == m) flag = 1;
      continue;
    }
    if (check(i + 1, a, x[i])) flag = 1;
    if (check(i + 1, a, y[i])) flag = 1;
    break;
  }
  if (flag) {
    puts("YES");
    return 0;
  }
  puts("NO");
}
