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 double eps = 1e-5;
const int maxn = 3e5 + 5;
struct node {
  int a, b;
  bool operator<(const node &oth) const {
    if (a != oth.a) return a < oth.a;
    return b < oth.b;
  }
  bool operator==(const node &oth) const { return a == oth.a && b == oth.b; }
} w[maxn];
int n, m;
bool judge(int ans) {
  int i = 1;
  int p1 = -1, p2 = -1;
  for (; i <= m; i++)
    if (w[i].a != ans && w[i].b != ans) {
      p1 = i;
      break;
    }
  i++;
  for (; i <= m; i++)
    if (w[i].a != ans && w[i].b != ans) {
      p2 = i;
      break;
    }
  if (p2 == -1) return true;
  int ans2 = -1;
  int A[10] = {0, w[p1].a, w[p1].b, w[p2].a, w[p2].b};
  sort(A + 1, A + 1 + 4);
  for (int i = 1; i <= 4; i++)
    if (A[i] == A[i - 1]) {
      ans2 = A[i];
      break;
    }
  if (ans2 == -1) return false;
  for (i = 1; i <= m; i++) {
    if (w[i].a != ans && w[i].b != ans)
      if (w[i].a != ans2 && w[i].b != ans2) return false;
  }
  return true;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) cin >> w[i].a >> w[i].b;
  sort(w + 1, w + 1 + m);
  m = unique(w + 1, w + 1 + m) - w - 1;
  if (judge(w[1].a) || judge(w[1].b))
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}
