#include <bits/stdc++.h>
using namespace std;
bool fa[30];
int d[30];
int main() {
  int n;
  scanf("%d", &n);
  memset(fa, 0, sizeof(fa));
  for (int i = 0; i < n; i++) scanf("%d", &d[i]);
  sort(d, d + n);
  for (int i = 0; i < n; i++) {
    int k = 1, j;
    for (j = i; k < d[i] && j >= 0; j--) {
      if ((!fa[j]) && d[j] + 2 <= d[i] && d[j] + k <= d[i]) {
        k += d[j];
        fa[j] = 1;
      }
    }
    if (j < 0 && k < d[i]) {
      printf("NO\n");
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    if (!fa[i]) {
      printf("NO\n");
      return 0;
    }
  }
  printf("YES\n");
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
bool fa[30];
int d[30];
int main() {
  int n;
  scanf("%d", &n);
  memset(fa, 0, sizeof(fa));
  for (int i = 0; i < n; i++) scanf("%d", &d[i]);
  sort(d, d + n);
  for (int i = 0; i < n; i++) {
    int k = 1, j;
    for (j = i; k < d[i] && j >= 0; j--) {
      if ((!fa[j]) && d[j] + 2 <= d[i] && d[j] + k <= d[i]) {
        k += d[j];
        fa[j] = 1;
      }
    }
    if (j < 0 && k < d[i]) {
      printf("NO\n");
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    if (!fa[i]) {
      printf("NO\n");
      return 0;
    }
  }
  printf("YES\n");
  return 0;
}

```