#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v.push_back(x);
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    if (v[i] == 1) continue;
    if (v[i] == 2) {
      puts("NO");
      return 0;
    }
    int req = v[i] - 1, num = 0;
    for (int j = i - 1; j >= 0; j--)
      if (v[j] < req or (v[j] == req and num > 0)) {
        req -= v[j];
        v.erase(v.begin() + j);
        j++;
        i--;
        n--;
        num++;
      }
    if (req != 0) {
      puts("NO");
      return 0;
    }
  }
  puts((int)v.size() == 1 ? "YES" : "NO");
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 main() {
  int n;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v.push_back(x);
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    if (v[i] == 1) continue;
    if (v[i] == 2) {
      puts("NO");
      return 0;
    }
    int req = v[i] - 1, num = 0;
    for (int j = i - 1; j >= 0; j--)
      if (v[j] < req or (v[j] == req and num > 0)) {
        req -= v[j];
        v.erase(v.begin() + j);
        j++;
        i--;
        n--;
        num++;
      }
    if (req != 0) {
      puts("NO");
      return 0;
    }
  }
  puts((int)v.size() == 1 ? "YES" : "NO");
  return 0;
}

```