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>
#pragma comment(linker, "/stack:20000000")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
constexpr int infValue =
    std::is_same<T, int>::value ? 2000100007 : 8600000000000000007ll;
template <typename T>
constexpr int maxValue =
    std::is_same<T, int>::value ? 1000000007 : 1000000000000000001ll;
const int INF = infValue<int>;
const long long MOD = 998244353ll;
const double EPS = 1e-6;
const int MAX = maxValue<int>;
using pr = pair<int, int>;
int n, m;
pr p[300010];
bool ok, dok = true;
void test(int x, int y) {
  dok = false;
  bool tok = true;
  for (int i = 0; i < m; ++i)
    tok &= x == p[i].first || x == p[i].second || y == p[i].second ||
           y == p[i].first;
  ok |= tok;
}
int main(void) {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; ++i) scanf("%d %d", &p[i].first, &p[i].second);
  sort(p, p + m);
  m = unique(p, p + m) - p;
  for (int i = 0; i < m; ++i) {
    if (p[i].first != p[0].first && p[i].second != p[0].first) {
      test(p[0].first, p[i].first);
      test(p[0].first, p[i].second);
      break;
    }
  }
  for (int i = 0; i < m; ++i) {
    if (p[i].first != p[0].second && p[i].second != p[0].second) {
      test(p[0].second, p[i].first);
      test(p[0].second, p[i].second);
      break;
    }
  }
  puts(ok || dok ? "YES" : "NO");
  return 0;
}
