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;
int main() {
  int n;
  cin >> n;
  if (n == 1)
    cout << "YES";
  else {
    int arr[n];
    for (int i = 0; i < n; i++) cin >> arr[i];
    int count = 0;
    int counte[n];
    int k = 0;
    for (int i = 0; i < n; i++) {
      if (arr[i] != -1) {
        count = 1;
        for (int j = i + 1; j < n; j++) {
          if (arr[j] == arr[i]) {
            count++;
            arr[j] = -1;
          }
        }
        counte[k] = count;
        k++;
        arr[i] = -1;
      }
    }
    if (k == 1)
      cout << "NO";
    else {
      for (int i = 0; i < k; i++) {
        for (int j = 0; j < k - i - 1; j++) {
          if (counte[j] > counte[j + 1]) {
            int temp = counte[j];
            counte[j] = counte[j + 1];
            counte[j + 1] = temp;
          }
        }
      }
      int d = counte[k - 1] - counte[k - 2];
      int sum = (k - 2) * counte[0];
      for (int i = 1; i < k - 1; i++) {
        sum = sum + (k - (i + 2)) * (counte[i] - counte[i - 1]);
      }
      if (sum + 1 >= d)
        cout << "YES";
      else
        cout << "NO";
    }
  }
  return 0;
}
