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.util.*;
import java.lang.*;

public class A {
    public static void main(String[] args) {
        // getting the inputs
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n];
        for(int i = 0; i < n; i++)
            a[i] = in.nextInt();
        // printing the output
        new Solver().solve(n, a);
    }
}

class Solver {
    public void solve(int n, int[] a) {  
        if(n == 1) {
            System.out.println("YES");
            return;
        }

        int i, j;
        HashSet<Integer> hs = new HashSet();
        for(i = 0; i < n; i++) {
            hs.add(a[i]);
        } 

        if(hs.size() < 2) {
            System.out.println("NO");
            return;
        }

        int[] count = new int[hs.size()];
        for(i = 0; i < hs.size(); i++) {
            count[i] = 0;
        }
        for(i = 0; i < hs.size(); i++) {
            int x =(int) new ArrayList(hs).get(i);
            for(j = 0; j < n; j++) {
                if(x == a[j]) count[i]++;
            }
        }
        if(n % 2 != 0) {
            for(i = 0; i < hs.size(); i++) {
                if(count[i] > (n/2 + 1)) {
                    System.out.println("NO");
                    return;
                }
            }
        }
        else {
            for(i = 0; i < hs.size(); i++) {
                if(count[i] > (n/2)) {
                    System.out.println("NO");
                    return;
                }
            }
        }
        System.out.println("YES");
    }
}