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 a[25], b[25], n;
int dfs(int x, int son, int have, int edge) {
  if (x == n) return (edge == n - 1);
  if (!son) return dfs(x + 1, a[x + 1] - 1, 0, edge);
  int st = (have ? son : son - 1);
  for (int i = 1; i <= st; i++) {
    if (b[i]) {
      b[i]--;
      if (dfs(x, son - i, have + 1, edge + 1)) {
        return 1;
      }
      b[i]++;
    }
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[a[i]]++;
    if (a[i] == 2) {
      printf("NO\n");
      return 0;
    }
  }
  sort(a, a + n);
  int ok = dfs(0, a[0] - 1, 0, 0);
  if (ok) {
    printf("YES\n");
  } else {
    printf("NO\n");
  }
}
