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 f1(int a[], int b[], int n) {
  int c, d, y = 0, z = 0, i;
  c = a[0];
  for (i = 0; i < n; i++) {
    if (c == a[i] || c == b[i]) {
      a[i] = 0;
      b[i] = 0;
    }
  }
  for (i = 0; i < n; i++) {
    if (a[i] != 0) {
      d = a[i];
      y = i;
      break;
    }
  }
  for (i = y; i < n; i++) {
    if (d == a[i] || d == b[i]) {
      a[i] = 0;
      b[i] = 0;
    }
  }
  for (i = 0; i < n; i++) {
    if (a[i] == 0 && b[i] == 0) z++;
  }
  return z;
}
int f2(int a[], int b[], int n) {
  int c, d, y = 0, z = 0, i;
  c = b[0];
  for (i = 0; i < n; i++) {
    if (c == a[i] || c == b[i]) {
      a[i] = 0;
      b[i] = 0;
    }
  }
  for (i = 0; i < n; i++) {
    if (b[i] != 0) {
      d = b[i];
      y = i;
      break;
    }
  }
  for (i = y; i < n; i++) {
    if (d == a[i] || d == b[i]) {
      a[i] = 0;
      b[i] = 0;
    }
  }
  for (i = 0; i < n; i++) {
    if (a[i] == 0 && b[i] == 0) z++;
  }
  return z;
}
int f3(int a[], int b[], int n) {
  int c, d, y = 0, z = 0, i;
  c = a[0];
  for (i = 0; i < n; i++) {
    if (c == a[i] || c == b[i]) {
      a[i] = 0;
      b[i] = 0;
    }
  }
  for (i = 0; i < n; i++) {
    if (a[i] != 0) {
      d = b[i];
      y = i;
      break;
    }
  }
  for (i = y; i < n; i++) {
    if (d == a[i] || d == b[i]) {
      a[i] = 0;
      b[i] = 0;
    }
  }
  for (i = 0; i < n; i++) {
    if (a[i] == 0 && b[i] == 0) z++;
  }
  return z;
}
int f4(int a[], int b[], int n) {
  int c, d, y = 0, z = 0, i;
  c = b[0];
  for (i = 0; i < n; i++) {
    if (c == a[i] || c == b[i]) {
      a[i] = 0;
      b[i] = 0;
    }
  }
  for (i = 0; i < n; i++) {
    if (a[i] != 0) {
      d = a[i];
      y = i;
      break;
    }
  }
  for (i = y; i < n; i++) {
    if (d == a[i] || d == b[i]) {
      a[i] = 0;
      b[i] = 0;
    }
  }
  for (i = 0; i < n; i++) {
    if (a[i] == 0 && b[i] == 0) z++;
  }
  return z;
}
int main() {
  int n, m, c, d, u, v, i;
  scanf("%d%d", &n, &m);
  int a[m], b[m], x[m], y[m];
  for (i = 0; i < m; i++) {
    scanf("%d%d", &a[i], &b[i]);
    x[i] = a[i];
    y[i] = b[i];
  }
  c = f1(x, y, m);
  for (i = 0; i < m; i++) {
    x[i] = a[i];
    y[i] = b[i];
  }
  d = f2(x, y, m);
  for (i = 0; i < m; i++) {
    x[i] = a[i];
    y[i] = b[i];
  }
  u = f3(x, y, m);
  for (i = 0; i < m; i++) {
    x[i] = a[i];
    y[i] = b[i];
  }
  v = f4(x, y, m);
  if (c == m || d == m || u == m || v == m)
    printf("YES");
  else
    printf("NO");
  return 0;
}
