#include <bits/stdc++.h>
using namespace std;
int n, A[30], vis[30];
int dfs(int node, int sum, int ch) {
  if (node == n + 1) return 1;
  if (A[node] - sum == 1) {
    if (ch == 1)
      return 0;
    else
      return dfs(node + 1, 0, 0);
  }
  int pre = -1;
  for (int i = 1; i <= n; i++) {
    if (A[i] != pre && !vis[i] && A[node] >= A[i] + sum) {
      pre = A[i];
      vis[i] = 1;
      if (dfs(node, sum + A[i], ch + 1)) return 1;
      vis[i] = 0;
    }
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &A[i]);
  }
  sort(A + 1, A + n + 1);
  if (A[n] != n)
    printf("NO");
  else {
    if (dfs(1, 0, 0))
      printf("YES");
    else
      printf("NO");
  }
  return 0;
}

### 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 n, A[30], vis[30];
int dfs(int node, int sum, int ch) {
  if (node == n + 1) return 1;
  if (A[node] - sum == 1) {
    if (ch == 1)
      return 0;
    else
      return dfs(node + 1, 0, 0);
  }
  int pre = -1;
  for (int i = 1; i <= n; i++) {
    if (A[i] != pre && !vis[i] && A[node] >= A[i] + sum) {
      pre = A[i];
      vis[i] = 1;
      if (dfs(node, sum + A[i], ch + 1)) return 1;
      vis[i] = 0;
    }
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &A[i]);
  }
  sort(A + 1, A + n + 1);
  if (A[n] != n)
    printf("NO");
  else {
    if (dfs(1, 0, 0))
      printf("YES");
    else
      printf("NO");
  }
  return 0;
}

```