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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
import java.util.StringTokenizer;


public class CodeForce179C {
	public static void main(String[] args) throws NumberFormatException,
			IOException {

		InputStream inputStream = System.in;
		BufferedReader in = new BufferedReader(new InputStreamReader(inputStream));
		StringTokenizer token;
		int n = Integer.parseInt(in.readLine());
		token = new StringTokenizer(in.readLine());
		int [] numbers= new int[n];
		for(int i=0;i<n;i++) {
			numbers[i] = Integer.parseInt(token.nextToken());
		}
		Map<Integer, Integer> count = new HashMap<Integer, Integer>();
		for(int i=0;i<n;i++) {
			if(count.containsKey(numbers[i])) {
				count.put(numbers[i], count.get(numbers[i])+1);
			}
			else {
				count.put(numbers[i], 1);
			}
		}
		int flag=0;
		for(Entry<Integer, Integer> eachCount: count.entrySet()) {
			if(eachCount.getValue()>((n/2)+n%2)) {
				flag=1;
				break;
			}
		}
		if(flag==1)
			System.out.println("NO");
		else
			System.out.println("YES");
 	}
}
