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.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class YaroslavAndPermutations {
	public static void main(String[] args) throws IOException {
	  Scanner sc = new Scanner(System.in);
		
	  //READ----------------------------------------------------
    int n = sc.nextInt();
    int[] t = new int[n];
    int[] sa = new int[1000+1];
    for (int i = 0; i < t.length; i++){
      t[i] = sc.nextInt();
      sa[t[i]]++;
     }
    
    Arrays.sort(sa);
		
    //SOLVE----------------------------------------------------
		int maxSame = sa[sa.length-1];
//		1, last = t[0], same = 1;
//		for (int i = 1; i < t.length; i++)
//    {
//      if(t[i]==last)
//        same++;
//      else{
//        if(same>maxSame)
//          maxSame=same;
//        last = t[i];
//        same = 1;
//      }
//    }
//		
//		if(same>maxSame)
//      maxSame=same;
//		
		if(n%2==1 && maxSame<=n/2+1)
  		System.out.println("YES");
		else if(n%2==0 && maxSame<=n/2)
      System.out.println("YES");
		else
		  System.out.println("NO");
    
    //CLOSE----------------------------------------------------
    sc.close();
	}
}