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, sn, ln, mn, a[25], b[1 << 12];
bool f[2][1 << 12][25];
bool get(int s, int i) {
  for (int su = s; su; su = s & (su - 1))
    if (!(su & (s ^ su)))
      for (int j = 0; j <= i; j++)
        if ((f[0][su][j] || f[1][su][j]) &&
            (f[0][su ^ s][i - j] || f[1][s ^ su][i - j]))
          return true;
  for (int j = 0; j <= i; j++)
    if ((f[0][0][j] || f[1][0][j]) && (f[0][s][i - j] || f[1][s][i - j]))
      return true;
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n, greater<int>());
  for (int i = 0; i < n; i++) mn += (a[i] > 1);
  ln = n - mn;
  sn = 1 << mn;
  sort(a, a + mn);
  if (mn * 2 >= n) return puts("NO"), 0;
  for (int i = 1; i <= sn; i++) b[i] = b[i >> 1] + (i & 1);
  f[0][0][1] = true;
  for (int s = 0; s < sn; s++)
    for (int i = 0; i <= ln; i++)
      if (get(s, i)) {
        f[1][s][i] = true;
        for (int t = 0; t < mn; t++)
          if (!(s & (1 << t)) && a[t] == b[s] + i + 1)
            f[0][s ^ (1 << t)][i] = true;
      }
  puts(f[0][sn - 1][ln] ? "YES" : "NO");
}
