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 els, aux;
  cin >> els;
  int all[els];
  bool diff = true;
  if (els == 1)
    cout << "YES";
  else {
    for (int i = 0; i < els; i++) cin >> all[i];
    for (int i = 0; i < els - 1; i++) {
      if (all[i] == all[i + 1]) diff = false;
    }
    if (diff)
      cout << "YES";
    else {
      for (int i = 0; i < els - 1; i++) {
        if (all[i] == all[i + 1] and i + 2 < els) {
          if (all[i] != all[i + 2]) {
            aux = all[i];
            all[i + 1] = all[i + 2];
            all[i + 2] = aux;
            i = -1;
          }
        }
      }
      diff = true;
      for (int i = 0; i < els - 1; i++) {
        if (all[i] == all[i + 1]) diff = false;
      }
      if (diff)
        cout << "YES";
      else {
        for (int i = els - 1; i > 0; i--) {
          if (all[i] == all[i - 1] and i - 2 >= 0) {
            if (all[i] != all[i - 2]) {
              aux = all[i];
              all[i - 1] = all[i - 2];
              all[i - 2] = aux;
              i = els;
            }
          }
        }
        diff = true;
        for (int i = 0; i < els - 1; i++) {
          if (all[i] == all[i + 1]) diff = false;
        }
        if (diff)
          cout << "YES";
        else
          cout << "NO";
      }
    }
  }
  return 0;
}
