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.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class YaroslavAndPermutations {
	
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		Solution sol = new Solution();
		sol.solve(in, out);
		out.close();
	}

	private static class Solution {
		private void solve(InputReader in, PrintWriter out) {
			int n=in.ni();
			int[] a=new int[1002];
			for (int i=0; i<n; ++i)
				a[in.ni()]++;
			for (int i=1; i<=1000; ++i)
				if (a[i]>(n+1)/2) {
					out.println("NO");
					return;
				}
			out.println("YES");
		}
	}

	private static class InputReader {
		private BufferedReader reader;
		private StringTokenizer tokenizer;

		private InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		private String n() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		private int ni() {
			return Integer.parseInt(n());
		}
	}
}
