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;
  long long M;
  cin >> N >> M;
  vector<pair<long long, long long>> v(M);
  for (long long i = 0; i < M; i++) {
    cin >> v[i].first >> v[i].second;
    if (v[i].second < v[i].first) swap(v[i].first, v[i].second);
  }
  sort(v.begin(), v.end());
  vector<pair<long long, long long>> ar;
  ar.push_back(v[0]);
  for (long long i = 1; i < M; i++) {
    if (v[i] != v[i - 1]) ar.push_back(v[i]);
  }
  long long X = ar[0].first;
  long long Y = ar[0].second;
  vector<pair<long long, long long>> d;
  vector<pair<long long, long long>> r;
  for (long long i = 0; i < ar.size(); i++) {
    if (ar[i].first != X && ar[i].second != X) d.push_back(ar[i]);
    if (ar[i].first != Y && ar[i].second != Y) r.push_back(ar[i]);
  }
  if (d.size() == 0 || r.size() == 0) {
    cout << "YES";
    return 0;
  }
  long long xx = d[0].first;
  long long yx = d[0].second;
  long long xy = r[0].first;
  long long yy = r[0].second;
  long long cx = 0, cy = 0, dx = 0, dy = 0;
  for (long long i = 0; i < d.size(); i++) {
    if (d[i].first != xx && d[i].second != xx) cx++;
    if (d[i].first != yx && d[i].second != yx) cy++;
  }
  for (long long i = 0; i < r.size(); i++) {
    if (r[i].first != xy && r[i].second != xy) dx++;
    if (r[i].first != yy && r[i].second != yy) dy++;
  }
  if (cx == 0 || cy == 0 || dx == 0 || dy == 0)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
