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[30], b[30], son[30], n;
bool solve(int dep) {
  if (dep == n) {
    for (int i = 0; i < n; ++i)
      if (a[i] > 1 || (b[i] > 1 && son[i] < 2)) return false;
    return true;
  }
  for (int i = 0; i < dep; ++i) {
    if (a[dep] < a[i] - (!son[i])) {
      a[i] -= a[dep];
      son[i]++;
      if (solve(dep + 1)) return true;
      a[i] += a[dep];
      son[i]--;
    }
  }
  return false;
}
int main() {
  scanf("%d", &n);
  bool flag = 1;
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    flag &= a[i] != 2;
  }
  sort(a, a + n);
  reverse(a, a + n);
  copy(a, a + n, b);
  flag &= (a[0] == n);
  fill(son, son + n, 0);
  if (flag && solve(1))
    puts("YES");
  else
    puts("NO");
  return 0;
}
