Yaroslav has an array that consists of n integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.

Help Yaroslav.

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the array. The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000) — the array elements.

Output

In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.

Examples

Input

1
1


Output

YES


Input

3
1 1 2


Output

YES


Input

4
7 7 7 7


Output

NO

Note

In the first sample the initial array fits well.

In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.

In the third sample Yarosav can't get the array he needs.
#include <bits/stdc++.h>
using namespace std;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<long long> primes;
bool prime[10005];
void seive() {
  memset(prime, 1, sizeof(prime));
  prime[0] = 0;
  prime[1] = 0;
  for (long long i = 2; i <= 10000; i++) {
    if (prime[i] == 1) {
      for (long long j = i * i; j <= 10000; j += i) prime[j] = 0;
    }
  }
}
long long power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans % 1000000007 * a % 1000000007) % 1000000007;
    a = (a * a) % 1000000007;
    b = b / 2;
  }
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n], i, j;
  vector<long long> v;
  map<long long, long long> m;
  map<long long, long long>::iterator it;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  long long ans = 0;
  for (it = m.begin(); it != m.end(); it++) {
    if (it->second > ans) ans = it->second;
  }
  if (n % 2 == 0) {
    if (ans > n / 2)
      cout << "NO";
    else
      cout << "YES";
  }
  if (n % 2 == 1) {
    n++;
    if (ans > n / 2)
      cout << "NO";
    else
      cout << "YES";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  bool codechef = 0;
  long long t = 1;
  if (codechef) cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
