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;
pair<int, int> arr[(int)1e5 * 3 + 1];
vector<pair<int, int> > v;
bool ny, main_ny = 0;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int x, y, n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) cin >> arr[i].first >> arr[i].second;
  x = arr[0].first;
  for (int i = 0; i < m; i++) {
    if (arr[i].first != x && arr[i].second != x) v.push_back(arr[i]);
  }
  if (!v.empty()) {
    int sz = v.size();
    y = v[0].first;
    ny = 1;
    for (int i = 0; i < sz; i++) {
      if (v[i].first != y && v[i].second != y) ny = 0;
    }
    if (ny) main_ny = 1;
    y = v[0].second;
    ny = 1;
    for (int i = 0; i < sz; i++) {
      if (v[i].first != y && v[i].second != y) ny = 0;
    }
    if (ny) main_ny = 1;
    v.clear();
  } else {
    cout << "YES\n";
    return 0;
  }
  x = arr[0].second;
  for (int i = 0; i < m; i++) {
    if (arr[i].first != x && arr[i].second != x) v.push_back(arr[i]);
  }
  if (!v.empty()) {
    int sz = v.size();
    y = v[0].first;
    ny = 1;
    for (int i = 0; i < sz; i++) {
      if (v[i].first != y && v[i].second != y) ny = 0;
    }
    if (ny) main_ny = 1;
    y = v[0].second;
    ny = 1;
    for (int i = 0; i < sz; i++) {
      if (v[i].first != y && v[i].second != y) ny = 0;
    }
    if (ny) main_ny = 1;
    v.clear();
  } else {
    cout << "YES\n";
    return 0;
  }
  if (main_ny)
    cout << "YES\n";
  else
    cout << "NO\n";
}
