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;
const int MAXN = 3e5 + 5;
pair<int, int> a[MAXN];
int vis[MAXN];
int num[MAXN];
int n, m;
vector<pair<int, int>> vec;
int ans = false;
signed main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    if (u > v) swap(u, v);
    a[i] = make_pair(u, v);
  }
  sort(a + 1, a + 1 + m);
  for (int i = 1; i <= m; ++i) {
    if (a[i].first == a[1].first || a[i].second == a[1].first) vis[i] = 1;
  }
  for (int i = 1; i <= m; ++i) {
    if (!vis[i]) vec.push_back(a[i]);
  }
  int cnt;
  cnt = vec.size();
  for (int i = 0; i < vec.size(); ++i) {
    num[vec[i].first]++;
    num[vec[i].second]++;
  }
  for (int i = 1; i <= n; ++i) {
    if (num[i] == cnt) ans = true;
  }
  memset(vis, 0, sizeof(vis));
  memset(num, 0, sizeof(num));
  vec.clear();
  for (int i = 1; i <= m; ++i) {
    if (a[i].second == a[1].second || a[i].first == a[1].second) vis[i] = 1;
  }
  for (int i = 1; i <= m; ++i) {
    if (!vis[i]) vec.push_back(a[i]);
  }
  cnt = vec.size();
  for (int i = 0; i < vec.size(); ++i) {
    num[vec[i].first]++;
    num[vec[i].second]++;
  }
  for (int i = 1; i <= n; ++i) {
    if (num[i] == cnt) ans = true;
  }
  if (ans)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
