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.*;

public class Problems
{

    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int[] a = new int[n];
        int[] distinct = new int[n];
        int i, i2, check, sum;
        for (i = 0; i < n; i++)
        {
            a[i] = input.nextInt();
            distinct[i] = 0;
        }
        if (n == 1)
        {
            System.out.println("YES");
        }
        else
        {
            for (i = 0; i < n; i++)
            {
                if (i == 0)
                {
                    distinct[0] = a[i];
                }
                else
                {
                    i2 = 0;
                    check = 0;
                    while (i2 < n && distinct[i2] != 0)
                    {
                        if (a[i] == distinct[i2])
                        {
                            check++;
                            break;
                        }
                        i2++;
                    }
                    if (check == 0)
                    {
                        distinct[i2] = a[i];
                    }
                }
            }
            i2 = 0;
            int x=0;
            while (i2 < n && distinct[i2] != 0)
            {
                sum = 0;
                for (i = 0; i < n; i++)
                {
                    if (distinct[i2] == a[i])
                    {
                        sum++;
                    }
                }
                if(x<sum)
                {
                    x=sum;
                }
                i2++;
            }
            if(n>=(x+(x-1)))
            {
                System.out.println("YES");
            }
            else
            {
                System.out.println("NO");
            }
        }
    }
}