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;
long long t, d, ct, q, ans, l, m, r, n, mx, mn, i, j, k, x, y, z, s;
char ch;
string ss = "", str = "";
map<pair<long long, long long>, long long> mp;
set<long long> st;
long long a[300003], b[300003];
vector<pair<long long, long long>> v;
bool ask(long long k) {
  long long i;
  for (i = 0; i < v.size(); i++)
    if (v[i].first != k && v[i].second != k) break;
  if (i == v.size()) return 1;
  long long x = v[i].first;
  long long g = k;
  if (x > k) swap(x, k);
  if (a[x] + a[k] - mp[{x, k}] == m) return 1;
  x = v[i].second;
  if (x > g) swap(x, g);
  if (a[x] + a[g] - mp[{x, g}] == m) return 1;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  s = k = d = l = mx = m = mn = r = q = i = j = z = x = y = ct = ans = 0;
  cin >> n >> m;
  for (i = 0; i < m; i++) {
    cin >> x >> y, a[x]++, a[y]++;
    v.push_back({x, y});
    if (x > y) swap(x, y);
    mp[{x, y}]++;
  }
  x = v[0].first;
  y = v[0].second;
  bool ya = ask(x);
  if (ya) {
    cout << "YES";
    return 0;
  }
  ya = ask(y);
  if (ya) {
    cout << "YES";
    return 0;
  }
  cout << "NO";
  return 0;
}
