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;
const int N = 33;
int a[N], b[N], c[N], n;
bool dfs(int k) {
  if (!k) {
    for (int i = 1; i <= n; ++i)
      if (b[i] != a[i] - 1) return 0;
    return 1;
  }
  for (int i = n; i > k; --i)
    if (a[i] - a[k] > 1 && a[k] + b[i] < a[i]) {
      b[i] += a[k];
      if (dfs(k - 1)) return 1;
      b[i] -= a[k];
    }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  sort(a + 1, a + 1 + n);
  if (n == 1) {
    if (a[1] == 1)
      puts("YES");
    else
      puts("NO");
  } else {
    for (int i = 1; i <= n; ++i)
      if (a[i] == 2) {
        puts("NO");
        return 0;
      }
    if (a[n] == n && dfs(n - 1))
      puts("YES");
    else
      puts("NO");
  }
}
