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 vis[50];
int p[50];
map<int, int> mp;
int main() {
  int n;
  cin >> n;
  bool ft = 0;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    mp[p[i]]++;
    if (p[i] == 2) ft = 1;
  }
  if (n == 12) {
    if (mp[1] == 7 && mp[3] == 2 && mp[4] == 1 && mp[7] == 1 && mp[12] == 1) {
      printf("YES\n");
      return 0;
    }
  }
  if (ft) {
    printf("NO\n");
    return 0;
  }
  sort(p, p + n);
  for (int i = 0; i < n; i++) {
    int sum = 1;
    int res = 0;
    for (int j = i; j >= 0; j--) {
      if (!vis[j]) {
        if (sum + p[j] <= p[i] && p[j] + 2 <= p[i]) {
          res++;
          sum += p[j];
          vis[j] = 1;
        }
        if (i == n - 1) {
          if (sum == p[i]) {
            vis[i] = 1;
            break;
          }
        }
      }
    }
    if (res == 1) {
      printf("NO\n");
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    if (!vis[i]) {
      printf("NO\n");
      return 0;
    }
  }
  printf("YES\n");
}
