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;
bool isvowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' ||
      c == 'E' || c == 'I' || c == 'O' || c == 'U')
    return true;
  return false;
}
long long fast_exp(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans *= x;
    n = n >> 1;
    x = x * x;
  }
  return ans;
}
long long mod_exp(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long sod(long long n) {
  long long sum = 0;
  while (n != 0) {
    sum = sum + (n % 10);
    n = n / 10;
  }
  return sum;
}
long long modInverse(long long a, long long m) { return mod_exp(a, m - 2, m); }
long long nCrModpDP(long long n, long long r, long long p) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, r); j > 0; j--) C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
long long nCrModpLucas(long long n, long long r, long long p) {
  if (r == 0) return 1;
  long long ni = n % p, ri = r % p;
  return (nCrModpLucas(n / p, r / p, p) * nCrModpDP(ni, ri, p)) % p;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<long long> a(m), b(m);
  for (long long i = (long long)0; i < (long long)m; i++) cin >> a[i] >> b[i];
  long long x, y;
  vector<long long> vec1;
  long long x1 = a[0];
  for (long long i = (long long)1; i < (long long)m; i++) {
    if (a[i] != x1 && b[i] != x1) vec1.push_back(i);
  }
  if (vec1.size() == 0) {
    cout << "YES\n";
    return;
  }
  long long y1 = a[vec1[0]];
  long long y2 = b[vec1[0]];
  bool foo = true;
  for (long long i = (long long)1; i < (long long)vec1.size(); i++) {
    if (a[vec1[i]] != y1 && b[vec1[i]] != y1) foo = false;
  }
  if (foo) {
    cout << "YES\n";
    return;
  }
  foo = true;
  for (long long i = (long long)1; i < (long long)vec1.size(); i++) {
    if (a[vec1[i]] != y2 && b[vec1[i]] != y2) foo = false;
  }
  if (foo) {
    cout << "YES\n";
    return;
  }
  vector<long long> vec2;
  long long x2 = b[0];
  for (long long i = (long long)1; i < (long long)m; i++) {
    if (a[i] != x2 && b[i] != x2) vec2.push_back(i);
  }
  if (vec2.size() == 0) {
    cout << "YES\n";
    return;
  }
  y1 = a[vec2[0]];
  y2 = b[vec2[0]];
  foo = true;
  for (long long i = (long long)1; i < (long long)vec2.size(); i++) {
    if (a[vec2[i]] != y1 && b[vec2[i]] != y1) foo = false;
  }
  if (foo) {
    cout << "YES\n";
    return;
  }
  foo = true;
  for (long long i = (long long)1; i < (long long)vec2.size(); i++) {
    if (a[vec2[i]] != y2 && b[vec2[i]] != y2) foo = false;
  }
  if (foo) {
    cout << "YES\n";
    return;
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  t = 1;
  while (t--) solve();
}
