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.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * 
 * 1 1 1 4 4
 * 1 1 2 2
 * 1 2 1 2
 * 1 1 2 1 2 1 2
 * 
 * */
public class A_Div2_67 {
    public static void main(String[]arg) throws IOException
    {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        /*long a = Integer.parseInt(in.readLine());
        long b = Integer.parseInt(in.readLine());
        long s = a+b,sz = 0,d,n,az=0,bz=0,pow=1;
        while(a > 0)
        {
            d = a%10;
            if(d != 0)
            {
                az = d*pow + az;
                pow *= 10;
            }
            a /= 10; 
        }
        pow = 1;
        while(b > 0)
        {
            d = b%10;
            if(d != 0)
            {
                bz = d*pow + bz;
                pow *= 10;
            }
            b /= 10; 
        }
        n = s;
        s = 0;
        pow = 1;
        while(n > 0)
        {
            d = n%10;
            if(d != 0)
            {
                s = d*pow + s;
                pow *= 10;
            }
            n/=10;
        }
        //System.out.println(az+"+"+bz+"="+s);
        if(az+bz == s)
            System.out.println("YES");
        else
            System.out.println("NO");*/
        int[]count = new int[1001];
        int i,n = Integer.parseInt(in.readLine());
        StringTokenizer st = new StringTokenizer(in.readLine());
        for(i = 0; i < n; i++)
        {
            count[Integer.parseInt(st.nextToken())]++;
        }
        Arrays.sort(count);
        int tam = (n%2) == 0 ? n/2 : (n/2) + 1;
        if(count[1000] <= tam)
        {
            System.out.println("YES");
        }
        else
            System.out.println("NO");
    }
}
