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
#include <bits/stdc++.h>
using namespace std;
int N, M, A[25];
bool V[25];
bool DFS(int p, int s, int k) {
  if (p > N) return M == N - 1;
  if (!k) return DFS(p + 1, p, A[p + 1]);
  for (; s; s--)
    if (!V[s] && A[s] + 1 < A[p] && A[s] + 1 <= k) {
      V[s] = 1;
      k -= A[s] + 1;
      M++;
      if (DFS(p, s - 1, k)) return 1;
      V[s] = 0;
      k += A[s] + 1;
      M--;
    }
  return 0;
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]), A[i]--;
  sort(A + 1, A + N + 1);
  puts(DFS(1, 0, A[1]) ? "YES" : "NO");
}
