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;
int N, M, A[300005], B[300005], x, y, ans = false, did = false;
vector<pair<int, int> > V;
bool chk[300005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;
  for (int i = 1; i <= M; i++) {
    cin >> A[i] >> B[i];
    if (A[i] > B[i]) swap(A[i], B[i]);
  }
  x = A[1];
  int idx = 0, cnt = 0, pos = 1;
  for (int i = 2; i <= M; i++) {
    if (x == A[i] || x == B[i])
      chk[i] = 1;
    else
      idx = i;
  }
  y = A[idx];
  for (int i = 2; i <= M; i++) {
    if (!chk[i]) {
      if (y == A[i] || y == B[i]) continue;
      pos = 0;
      break;
    }
  }
  if (pos) ans = 1;
  pos = 1;
  y = B[idx];
  for (int i = 2; i <= M; i++) {
    if (!chk[i]) {
      if (y == A[i] || y == B[i]) continue;
      pos = 0;
      break;
    }
  }
  if (pos) ans = 1;
  x = B[1];
  idx = 0, cnt = 0, pos = 1;
  memset(chk, 0, sizeof(chk));
  for (int i = 2; i <= M; i++) {
    if (x == A[i] || x == B[i])
      chk[i] = 1;
    else
      idx = i;
  }
  y = A[idx];
  for (int i = 2; i <= M; i++) {
    if (!chk[i]) {
      if (y == A[i] || y == B[i]) continue;
      pos = 0;
      break;
    }
  }
  if (pos) ans = 1;
  pos = 1;
  y = B[idx];
  for (int i = 2; i <= M; i++) {
    if (!chk[i]) {
      if (y == A[i] || y == B[i]) continue;
      pos = 0;
      break;
    }
  }
  if (pos) ans = 1;
  if (ans)
    cout << "YES";
  else
    cout << "NO";
}
