#include <bits/stdc++.h>
using namespace std;
const int MaxN = 30;
int u[MaxN], c[MaxN];
int main(void) {
  int n;
  while (scanf("%d", &n) != EOF) {
    int i, j;
    for (i = 0; i < n; i++) scanf(" %d", &c[i]);
    sort(c, c + n);
    reverse(c, c + n);
    if (c[0] != n || c[n / 2] != 1) {
      printf("NO\n");
      continue;
    }
    int ok = 1;
    memset(u, 0, sizeof(u));
    for (i = 0; i < n; i++) {
      int sum = 1, k = 0;
      for (j = i + 1; j < n && sum < c[i]; j++)
        if (u[j] == 0)
          if (sum + c[j] <= c[i] && !(k == 0 && sum + c[j] == c[i]))
            u[j] = 1, sum += c[j], k++;
      if (sum != c[i]) ok = 0;
    }
    if (!ok)
      printf("NO\n");
    else
      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;
const int MaxN = 30;
int u[MaxN], c[MaxN];
int main(void) {
  int n;
  while (scanf("%d", &n) != EOF) {
    int i, j;
    for (i = 0; i < n; i++) scanf(" %d", &c[i]);
    sort(c, c + n);
    reverse(c, c + n);
    if (c[0] != n || c[n / 2] != 1) {
      printf("NO\n");
      continue;
    }
    int ok = 1;
    memset(u, 0, sizeof(u));
    for (i = 0; i < n; i++) {
      int sum = 1, k = 0;
      for (j = i + 1; j < n && sum < c[i]; j++)
        if (u[j] == 0)
          if (sum + c[j] <= c[i] && !(k == 0 && sum + c[j] == c[i]))
            u[j] = 1, sum += c[j], k++;
      if (sum != c[i]) ok = 0;
    }
    if (!ok)
      printf("NO\n");
    else
      printf("YES\n");
  }
  return 0;
}

```