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;
string S;
struct point {
  int x, y;
} p[300005];
struct pair1 {
  int key, value;
  bool operator<(pair1 pp) const { return value > pp.value; }
} counter[300005];
vector<point> remain;
vector<int> cand;
bool solution1() {
  for (int i = 0; i < N + 1; i++) counter[i].key = i;
  for (int i = 0; i < M; i++) {
    counter[p[i].x].value++, counter[p[i].y].value++;
  }
  sort(counter, counter + N + 1);
  for (int i = 0; i < 4; i++) {
    if (counter[i].value < M / 2) return 0;
    int m = counter[i].key;
    if (0) cout << "m=" << m << endl;
    remain.clear(), cand.clear();
    for (int j = 0; j < M; j++) {
      if (p[j].x != m && p[j].y != m) remain.push_back(p[j]);
    }
    if (remain.empty()) return 1;
    cand.push_back(remain[0].x), cand.push_back(remain[0].y);
    for (auto e : remain) {
      if (cand.size() > 1 && cand[1] != e.x && cand[1] != e.y)
        cand.erase(cand.begin() + 1);
      if (cand[0] != e.x && cand[0] != e.y) cand.erase(cand.begin());
      if (0) {
        for (auto sss : cand) cout << sss << " ";
        cout << endl;
      }
      if (cand.empty()) goto NEXT;
    }
    return 1;
  NEXT:;
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M;
  for (int i = 0; i < M; i++) cin >> p[i].x >> p[i].y;
  cout << (solution1() ? "YES" : "NO");
  return 0;
}
