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 main() {
  long long n, m;
  cin >> n >> m;
  pair<long long, long long> p[m];
  long long toe[2] = {-1, -1};
  for (long long i = 0; i < m; i++) {
    cin >> p[i].first >> p[i].second;
    if (i > 0) {
      if (p[i].first != p[0].first && p[i].first != p[0].second &&
          p[i].second != p[0].first && p[i].second != p[0].second) {
        toe[0] = p[i].first;
        toe[1] = p[i].second;
      }
    }
  }
  if (toe[0] == -1) {
    cout << "YES" << endl;
    return 0;
  }
  long long a = p[0].first;
  long long b = toe[0];
  bool a1 = true;
  for (long long i = 0; i < m; i++) {
    if (p[i].first == a || p[i].second == a || p[i].first == b ||
        p[i].second == b) {
      ;
    } else {
      a1 = false;
      break;
    }
  }
  long long c = toe[1];
  bool a2 = true;
  for (long long i = 0; i < m; i++) {
    if (p[i].first == a || p[i].second == a || p[i].first == c ||
        p[i].second == c) {
      ;
    } else {
      a2 = false;
      break;
    }
  }
  a = p[0].second;
  bool a3 = true;
  for (long long i = 0; i < m; i++) {
    if (p[i].first == a || p[i].second == a || p[i].first == c ||
        p[i].second == c) {
      ;
    } else {
      a3 = false;
      break;
    }
  }
  bool a4 = true;
  for (long long i = 0; i < m; i++) {
    if (p[i].first == a || p[i].second == a || p[i].first == b ||
        p[i].second == b) {
      ;
    } else {
      a4 = false;
      break;
    }
  }
  if (a1 || a2 || a3 || a4)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
