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;
template <typename T>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while ('0' > ch || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
const int maxn = (3e5) + 10;
int n, m, c[maxn];
int x[maxn], y[maxn];
bool ck(int a, int b) {
  for (int i = 1; i <= m; i++)
    if (x[i] != a && x[i] != b && y[i] != a && y[i] != b) return 0;
  return 1;
}
int in[maxn], cnt, mx;
vector<int> g[maxn];
map<long long, bool> t;
bool ck(int a) {
  memset(in, 0, sizeof(in));
  cnt = mx = 0;
  for (int i = 1; i <= m; i++)
    if (x[i] != a && y[i] != a) in[x[i]]++, in[y[i]]++, cnt++;
  for (int i = 1; i <= n; i++) mx = max(mx, in[i]);
  return mx == cnt;
}
long long haxi(int x, int y) { return (long long)(x - 1) * n + y; }
int main() {
  read(n);
  read(m);
  long long tmp;
  for (int i = 1; i <= m; i++) {
    read(x[i]);
    read(y[i]);
    if (x[i] > y[i]) swap(x[i], y[i]);
    tmp = haxi(x[i], y[i]);
    if (t[tmp]) continue;
    t[tmp] = 1;
    g[x[i]].push_back(y[i]);
    g[y[i]].push_back(x[i]);
  }
  int a = 0, b = 0;
  for (int i = 1; i <= n; i++) {
    if (g[i].size() > 2) {
      if (!a)
        a = i;
      else if (!b) {
        b = i;
        break;
      } else {
        printf("No\n");
        return 0;
      }
    } else if (g[i].size() == 1) {
      if (a && ck(g[i][0], a)) {
        printf("Yes\n");
        return 0;
      }
      if (ck(g[i][0])) {
        printf("Yes\n");
        return 0;
      }
      if (!a)
        a = i;
      else if (!b) {
        b = i;
        break;
      } else {
        printf("No\n");
        return 0;
      }
    } else if (g[i].size() == 2) {
      if (ck(g[i][0], g[i][1])) {
        printf("Yes\n");
        return 0;
      }
      if (!a)
        a = i;
      else if (!b) {
        b = i;
        break;
      } else {
        printf("No\n");
        return 0;
      }
    }
  }
  if (ck(a, b))
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
