Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food. 

Iahub asks Iahubina: can you build a rooted tree, such that

  * each internal node (a node with at least one son) has at least two sons; 
  * node i has ci nodes in its subtree? 



Iahubina has to guess the tree. Being a smart girl, she realized that it's possible no tree can follow Iahub's restrictions. In this way, Iahub will eat all the food. You need to help Iahubina: determine if there's at least one tree following Iahub's restrictions. The required tree must contain n nodes.

Input

The first line of the input contains integer n (1 ≤ n ≤ 24). Next line contains n positive integers: the i-th number represents ci (1 ≤ ci ≤ n).

Output

Output on the first line "YES" (without quotes) if there exist at least one tree following Iahub's restrictions, otherwise output "NO" (without quotes). 

Examples

Input

4
1 1 1 4


Output

YES

Input

5
1 1 5 2 1


Output

NO
import java.util.Arrays;
import java.util.Scanner;


public class Main
{
	int n, a[], nextDif[];
	
	public boolean go ( int mask, int i, int sum, int j, int childCnt )
	{
		if ( i == n )
			return Integer.bitCount(mask) == 1;
		
		if ( sum+1 == a[i] ) {
			if ( childCnt < 2 ) return false;
			return go ( mask|(1<<i), i+1, 0, 0, 0 );
		}
		
		if ( j >= i ) return false;
		
		if ( ((mask>>j)&1) == 0 )
			return go ( mask, i, sum, j+1, childCnt );
		
		if ( a[j] + sum < a[i] )
			if ( go ( mask^(1<<j), i, sum+a[j], j+1, childCnt+1 ) )
				return true;
	
		if ( go ( mask, i, sum, nextDif[j], childCnt ) )
			return true;
		
		return false;
	}
	
	public void solve ( )
	{
		Scanner in = new Scanner ( System.in );
		n = in.nextInt();
		a = new int[n];
		for ( int i = 0; i < n; ++i )
			a[i] = in.nextInt();
		Arrays.sort(a);
		
		int init = 0;
		while ( init < n && a[init] == 1 ) init++;
		
		nextDif = new int[n];
		nextDif[n-1] = n;
		for ( int i = n-2; i >= 0; --i )
			if ( a[i] == a[i+1] ) nextDif[i] = nextDif[i+1];
			else nextDif[i] = i+1;
		
		String ans = "NO";
		if ( init > 0 )
			if ( go ( (1<<init)-1, init, 0, 0, 0 ) )
				ans = "YES";
		
		System.out.println(ans);
	}
	
	public static void main(String[] args) throws Exception {
		new Main().solve();
	}
}
