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, m, K;
vector<int> deg;
int cnt[1 << 11], sum[1 << 11];
bool f[15][25][1 << 11];
void work() {
  int i, j, k, l, x;
  scanf("%d", &n);
  for (i = 0; i < n; ++i) {
    scanf("%d", &x);
    if (x == 1)
      ++K;
    else
      deg.push_back(x);
  }
  m = deg.size();
  if (n == 1) {
    puts(m ? "NO" : "YES");
    return;
  }
  if (m > 11) {
    puts("NO");
    return;
  }
  sort(deg.begin(), deg.end());
  for (i = 0; i < 1 << m; ++i)
    for (j = 0; j < m; ++j)
      if (i >> j & 1) cnt[i] += deg[j], ++sum[i];
  f[0][0][0] = 1;
  for (i = 0; i < m; ++i)
    for (j = 0; j <= K; ++j)
      for (k = 0; k < 1 << m; ++k) {
        if (!f[i][j][k]) continue;
        x = ((1 << m) - 1) ^ k;
        for (l = x;; l = (l - 1) & x) {
          if (cnt[l] >= deg[i]) continue;
          if (deg[i] - cnt[l] - 1 <= K - j && sum[l] + deg[i] - cnt[l] > 2)
            f[i + 1][j + deg[i] - cnt[l] - 1][k | l] = 1;
          if (!l) break;
        }
      }
  puts(f[m][K][(1 << m - 1) - 1] ? "YES" : "NO");
}
int main() {
  work();
  return 0;
}
