#include <bits/stdc++.h>
using namespace std;
int n, p[24], w[24];
inline bool solve(int d) {
  if (d == n) return *max_element(p, p + n) == 0;
  int used = 0;
  for (int i = 0; i < (d); ++i) {
    if ((p[i] > w[d] || p[i] == w[d] && p[i] + 1 != w[i]) &&
        ((used >> p[i]) & 1) == 0) {
      p[i] -= w[d];
      if (solve(d + 1)) return true;
      p[i] += w[d];
      used |= 1 << p[i];
    }
  }
  return false;
}
int main() {
  cin >> n;
  for (int i = 0; i < (n); ++i) cin >> w[i];
  sort(w, w + n, greater<int>());
  for (int i = 0; i < (n); ++i) p[i] = w[i] - 1;
  cout << (solve(1) ? "YES" : "NO") << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, p[24], w[24];
inline bool solve(int d) {
  if (d == n) return *max_element(p, p + n) == 0;
  int used = 0;
  for (int i = 0; i < (d); ++i) {
    if ((p[i] > w[d] || p[i] == w[d] && p[i] + 1 != w[i]) &&
        ((used >> p[i]) & 1) == 0) {
      p[i] -= w[d];
      if (solve(d + 1)) return true;
      p[i] += w[d];
      used |= 1 << p[i];
    }
  }
  return false;
}
int main() {
  cin >> n;
  for (int i = 0; i < (n); ++i) cin >> w[i];
  sort(w, w + n, greater<int>());
  for (int i = 0; i < (n); ++i) p[i] = w[i] - 1;
  cout << (solve(1) ? "YES" : "NO") << endl;
  return 0;
}

```