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[28];
deque<unsigned long> sums[28];
bool used[28];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  unsigned long ma = 1ul << n;
  for (unsigned long i = 1; i < ma; i++)
    if (__builtin_popcount(i) > 1) {
      int sum = 0;
      for (int j = 0; j < n; j++) {
        sum += a[j] * ((i & (1ul << j)) ? 1 : 0);
      }
      if (sum && sum <= n - 1) sums[sum].push_front(i);
    }
  bool fullok = true;
  for (int i = 0; i < n && fullok; i++) {
    if (a[i] != 1) {
      fullok = false;
      int k = a[i] - 1;
      for (int it = 0; it < sums[k].size(); it++) {
        bool ok = true;
        int sum = 0;
        for (int j = 0; j < i && ok && sum != k; j++) {
          if (sums[k][it] & 1ul << j) {
            ok &= !used[j];
            sum += a[j];
          }
        }
        if (ok && sum == k) {
          for (int j = 0; j < i && ok; j++)
            if (sums[k][it] & 1ul << j) {
              used[j] = true;
            }
          sums[k].erase(sums[k].begin() + it);
          fullok = true;
          break;
        } else if (!ok) {
          it--;
          sums[k].erase(sums[k].begin() + it + 1);
        }
      }
    }
  }
  for (int i = 0; i < n - 1 && fullok; i++) {
    fullok &= used[i];
  }
  if (fullok) {
    cout << "YES";
  } else
    cout << "NO";
  return 0;
}
