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;
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
bool comparator(int i, int j) { return (i < j); }
long long gcd(long long x, long long y) {
  long long t;
  while (y != 0) {
    t = x % y;
    x = y;
    y = t;
  }
  return x;
}
bool isVowel(char ch) {
  if (ch == 'a' || ch == 'i' || ch == 'e' || ch == 'u' || ch == 'o' ||
      ch == 'y') {
    return true;
  }
  return false;
}
bool isPrime(int n) {
  int root = sqrt(n);
  for (int i = 2; i <= root; i++)
    if (n % i == 0) return 0;
  return 1;
}
int n, m;
pair<int, int> c[300004];
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) cin >> c[i].first >> c[i].second;
  int x = c[0].first;
  int cnt[300004] = {
      0,
  };
  int ans = 0;
  for (int i = 1; i < m; i++) {
    if (c[i].first == x || c[i].second == x)
      continue;
    else {
      cnt[c[i].first]++;
      cnt[c[i].second]++;
      ans++;
    }
  }
  for (int i = 0; i <= 300000; i++)
    if (cnt[i] == ans) {
      cout << "YES";
      return 0;
    }
  memset(cnt, 0, sizeof(cnt));
  ans = 0;
  x = c[0].second;
  for (int i = 1; i < m; i++) {
    if (c[i].first == x || c[i].second == x)
      continue;
    else {
      cnt[c[i].first]++;
      cnt[c[i].second]++;
      ans++;
    }
  }
  for (int i = 0; i <= 300000; i++)
    if (cnt[i] == ans) {
      cout << "YES";
      return 0;
    }
  cout << "NO";
  return 0;
}
