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 a[4000005];
struct E {
  int x, y;
};
E no[4000005];
int cmp(int c, int d) { return c > d; }
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  int x, y;
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &x, &y);
    no[i].x = x;
    no[i].y = y;
  }
  queue<E> q;
  for (int i = 1; i < m; i++) {
    if (no[i].x != no[0].x && no[i].y != no[0].x) {
      q.push(no[i]);
    }
  }
  int size = 0;
  int nn = q.size();
  while (!q.empty()) {
    E t = q.front();
    q.pop();
    if (a[t.x] == 0) size = max(t.x, size);
    if (a[t.y] == 0) size = max(t.y, size);
    a[t.x]++;
    a[t.y]++;
  }
  sort(a, a + size + 1, cmp);
  int ans = 0;
  if (a[0] >= nn) {
    printf("YES");
    return 0;
  } else {
    queue<E> b;
    memset(a, 0, sizeof(a));
    for (int i = 1; i < m; i++) {
      if (no[i].x != no[0].y && no[i].y != no[0].y) {
        b.push(no[i]);
      }
    }
    int size1 = 0;
    int nnb = b.size();
    while (!b.empty()) {
      E t = b.front();
      b.pop();
      if (a[t.x] == 0) size1 = max(size1, t.x);
      if (a[t.y] == 0) size1 = max(size1, t.y);
      a[t.x]++;
      a[t.y]++;
    }
    sort(a, a + size1 + 1, cmp);
    if (a[0] >= nnb) {
      printf("YES");
    } else
      printf("NO");
  }
  return 0;
}
