#include <bits/stdc++.h>
using namespace std;
int a[30], h[30], s[30];
void solution() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    h[i] = 1;
    s[i] = 0;
  }
  sort(&a[1], &a[n + 1]);
  for (int i = n - 1; i >= 1; i--) {
    int t = i + 1;
    while (h[t] + a[i] > a[t] || (h[t] + a[i] == a[t] && s[t] == 0)) t++;
    if (t > n) {
      cout << "NO" << '\n';
      ;
      return;
    }
    h[t] += a[i];
    s[t]++;
  }
  for (int i = 1; i <= n; i++)
    if (h[i] != a[i] || s[i] == 1) {
      cout << "NO" << '\n';
      return;
    }
  cout << "YES" << '\n';
}
int main() { solution(); }

### Prompt

Create a solution in CPP for 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 a[30], h[30], s[30];
void solution() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    h[i] = 1;
    s[i] = 0;
  }
  sort(&a[1], &a[n + 1]);
  for (int i = n - 1; i >= 1; i--) {
    int t = i + 1;
    while (h[t] + a[i] > a[t] || (h[t] + a[i] == a[t] && s[t] == 0)) t++;
    if (t > n) {
      cout << "NO" << '\n';
      ;
      return;
    }
    h[t] += a[i];
    s[t]++;
  }
  for (int i = 1; i <= n; i++)
    if (h[i] != a[i] || s[i] == 1) {
      cout << "NO" << '\n';
      return;
    }
  cout << "YES" << '\n';
}
int main() { solution(); }

```