#include <bits/stdc++.h>
const int N = 26;
using namespace std;
int n;
int a[N];
int d[N][N];
bool used[N];
void f() {
  d[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      d[i][j] = d[i - 1][j];
      if (j >= a[i] && !used[i]) d[i][j] |= d[i - 1][j - a[i]];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  if (a[n] != n) {
    cout << "NO" << endl;
    return 0;
  }
  f();
  for (int i = 1; i <= n; i++) {
    int need = a[i] - 1;
    for (int j = i - 1; j >= 1; j--) {
      if (need >= a[j] && d[j - 1][need - a[j]] && !used[j] &&
          a[j] != a[i] - 1) {
        used[j] = true;
        need -= a[j];
        f();
      }
    }
    if (need) {
      cout << "NO" << endl;
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    if (!used[i]) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  return 0;
}

### Prompt

Generate a Cpp solution 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>
const int N = 26;
using namespace std;
int n;
int a[N];
int d[N][N];
bool used[N];
void f() {
  d[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      d[i][j] = d[i - 1][j];
      if (j >= a[i] && !used[i]) d[i][j] |= d[i - 1][j - a[i]];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  if (a[n] != n) {
    cout << "NO" << endl;
    return 0;
  }
  f();
  for (int i = 1; i <= n; i++) {
    int need = a[i] - 1;
    for (int j = i - 1; j >= 1; j--) {
      if (need >= a[j] && d[j - 1][need - a[j]] && !used[j] &&
          a[j] != a[i] - 1) {
        used[j] = true;
        need -= a[j];
        f();
      }
    }
    if (need) {
      cout << "NO" << endl;
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    if (!used[i]) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  return 0;
}

```