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 cnt[25], f[25], t[25], n;
bool dfs(int step, int lim) {
  if (step == n) {
    for (int i = 0; i < n; ++i)
      if (f[i] != t[i]) return false;
    return true;
  }
  while (cnt[lim] == 0) lim--;
  f[step] = 1;
  t[step] = lim;
  for (int i = 0; i < step; ++i) {
    if (lim + 1 < t[i] && f[i] + lim <= t[i]) {
      f[i] += lim;
      cnt[lim]--;
      if (dfs(step + 1, lim)) return true;
      f[i] -= lim;
      cnt[lim]++;
      if (lim == 1) break;
    }
  }
  return false;
}
int main() {
  scanf("%d", &n);
  int tmp;
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < n; ++i) {
    scanf("%d", &tmp);
    cnt[tmp]++;
  }
  f[0] = 1;
  t[0] = n;
  if (cnt[2] || cnt[n] != 1)
    puts("NO");
  else if (!dfs(1, n - 1))
    puts("NO");
  else
    puts("YES");
  return 0;
}
