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.

import java.io.BufferedInputStream;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

/**
 * @author Evgeniy Kiselev
 *         Date: 11.04.13
 */
public class A {
    public static void main(String[] arg) {
        final Scanner scanner = new Scanner(new BufferedInputStream(System.in));
        final int size = Integer.parseInt(scanner.nextLine());
        final String[] input = scanner.nextLine().split(" ");
        if (size == 1) {
            System.out.println("YES");
            return;
        }
//        int[] arr = new int[size];
        Map<Integer, Integer> map = new HashMap<Integer, Integer>(size * 2);
        for (int i = 0; i < input.length; i++) {
//            arr[i] = Integer.parseInt(input[i]);
            int num = Integer.parseInt(input[i]);
            if (map.containsKey(num)) {
                map.put(num, map.get(num) + 1);
            } else {
                map.put(num, 1);
            }
        }
        int maxOccurence = 0;
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            maxOccurence = Math.max(maxOccurence, entry.getValue());
        }

        System.out.println(maxOccurence <= (size + (size % 2)) / 2 ? "YES" : "NO");
    }
}
