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>
const long long mod = 1000000000 + 7;
const long long INF = 9000000000000000000;
using namespace std;
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
long long gcd(long long a, long long b) { return (a == 0 ? b : gcd(b % a, a)); }
long long lcm(long long a, long long b) { return (a * b / (gcd(a, b))); }
long long add(long long a, long long b, long long c = 0) {
  return ((a % mod + b % mod) % mod + c % mod) % mod;
}
long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long sub(long long a, long long b) {
  return (a % mod - b % mod + mod) % mod;
}
void print(vector<long long> ans) {
  for (auto x : ans) cout << x << " ";
  cout << endl;
}
long long n, a, w, k, b, m, mx, cc, timer, realans, start, queries, startindex;
string s, t;
void solve() {
  cin >> n >> m;
  vector<pair<long long, long long> > arr;
  for (long long i = 0; i < m; i++) {
    cin >> a >> b;
    arr.push_back({min(a, b), max(a, b)});
  }
  if (arr.size() <= 2) {
    yes();
    return;
  }
  vector<long long> nums;
  nums.push_back(arr[0].first), nums.push_back(arr[0].second);
  for (auto x : nums) {
    vector<long long> cnt(n + 1, 0);
    long long all = 0;
    for (auto y : arr) {
      if (x != y.first && x != y.second) {
        cnt[y.first]++, cnt[y.second]++, all++;
      }
    }
    if (*max_element(cnt.begin(), cnt.end()) == all) {
      yes();
      return;
    }
  }
  no();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long tests = 1;
  while (tests--) {
    solve();
  }
  return 0;
}
