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 N, M;
pair<int, int> arr[300001];
bool visi[300001];
bool f(int x, int y) {
  bool done = true;
  for (int i = 1; i <= M; i++) {
    if (arr[i].first == x || arr[i].second == x || arr[i].first == y ||
        arr[i].second == y)
      continue;
    return false;
  }
  return true;
}
int main() {
  scanf("%d %d", &N, &M);
  for (int i = 1; i <= M; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    arr[i] = pair<int, int>(min(a, b), max(a, b));
  }
  pair<int, int> a = arr[1], b = pair<int, int>(-1, -1);
  for (int i = 2; i <= M; i++) {
    if (arr[i].first == a.first || arr[i].second == a.first ||
        arr[i].first == a.second || arr[i].second == a.second)
      continue;
    b = arr[i];
    break;
  }
  if (b.first == -1 || f(a.first, b.first) || f(a.first, b.second) ||
      f(a.second, b.first) || f(a.second, b.second)) {
    printf("YES");
  } else {
    printf("NO");
  }
}
