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;
void read(long long &a) {
  a = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    a = (a << 1) + (a << 3) + c - '0';
    c = getchar();
  }
}
long long Set(long long n, long long pos) { return n = n | (1LL << pos); }
long long reset(long long n, long long pos) { return n = n & ~(1LL << pos); }
bool check(long long n, long long pos) { return (bool)(n & (1LL << pos)); }
long long msb(long long mask) { return (mask & (-mask)); }
long long cnt_on(long long mask) { return __builtin_popcountll(mask); }
long long mul(long long a, long long b) {
  long long c;
  c = (a % 1000000007 * b % 1000000007) % 1000000007;
  return c;
}
long long add(long long a, long long b) {
  long long c;
  c = (a % 1000000007 + b % 1000000007) % 1000000007;
  return c;
}
long long sub(long long a, long long b) {
  long long c;
  c = ((a % 1000000007 - b % 1000000007) % 1000000007 + 1000000007) %
      1000000007;
  return c;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long exgcd(long long x, long long y, long long &a, long long &b) {
  int flag = 0;
  long long t, la = 1, lb = 0, ra = 0, rb = 1;
  while (x % y) {
    if (flag == 0)
      la -= x / y * ra, lb -= x / y * rb;
    else
      ra -= x / y * la, rb -= x / y * lb;
    t = x, x = y, y = t % y;
    flag = 1 - flag;
  }
  if (flag == 0)
    a = ra, b = rb;
  else
    a = la, b = lb;
  return y;
}
long long n, m;
vector<pair<long long, long long> > as;
long long cnt[300005];
int main() {
  scanf("%lld %lld", &n, &m);
  for (long long i = 1; i <= m; i++) {
    long long u, v;
    scanf("%lld %lld", &u, &v);
    if (u > v) swap(u, v);
    as.push_back(pair<long long, long long>(u, v));
  }
  sort(as.begin(), as.end());
  as.erase(unique(as.begin(), as.end()), as.end());
  if (as.size() == 1) {
    printf("YES\n");
    return 0;
  }
  long long x = as[0].first;
  long long cntt = 0;
  for (long long i = 1; i < as.size(); i++) {
    long long a = as[i].first;
    long long b = as[i].second;
    if (a == x || b == x) continue;
    cntt++;
    cnt[a]++;
    cnt[b]++;
  }
  for (long long i = 1; i <= n; i++) {
    if (cnt[i] == cntt) {
      printf("YES\n");
      return 0;
    }
  }
  x = as[0].second;
  memset((cnt), (0), sizeof(cnt));
  cntt = 0;
  for (long long i = 1; i < as.size(); i++) {
    long long a = as[i].first;
    long long b = as[i].second;
    if (a == x || b == x) continue;
    cntt++;
    cnt[a]++;
    cnt[b]++;
  }
  for (long long i = 1; i <= n; i++) {
    if (cnt[i] == cntt) {
      printf("YES\n");
      return 0;
    }
  }
  printf("NO\n");
  return 0;
}
