#include <bits/stdc++.h>
using namespace std;
int n;
int nodes[24];
int vis[24];
int dodo(int x, int cur, int child) {
  if (x == n) return 1;
  if (nodes[x] - cur == 1)
    if (child == 1)
      return 0;
    else
      return dodo(x + 1, 0, 0);
  int pre = -1;
  for (int i = 0; i < n; i++)
    if (nodes[i] != pre)
      if (!vis[i] && nodes[x] >= nodes[i] + cur) {
        pre = nodes[i];
        vis[i] = 1;
        if (dodo(x, cur + nodes[i], child + 1)) return 1;
        vis[i] = 0;
      }
  return 0;
}
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", nodes + i);
  sort(nodes, nodes + n);
  if (nodes[n - 1] != n)
    printf("NO\n");
  else
    dodo(0, 0, 0) ? printf("YES\n") : printf("NO\n");
}

### Prompt

Your task is to create 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>
using namespace std;
int n;
int nodes[24];
int vis[24];
int dodo(int x, int cur, int child) {
  if (x == n) return 1;
  if (nodes[x] - cur == 1)
    if (child == 1)
      return 0;
    else
      return dodo(x + 1, 0, 0);
  int pre = -1;
  for (int i = 0; i < n; i++)
    if (nodes[i] != pre)
      if (!vis[i] && nodes[x] >= nodes[i] + cur) {
        pre = nodes[i];
        vis[i] = 1;
        if (dodo(x, cur + nodes[i], child + 1)) return 1;
        vis[i] = 0;
      }
  return 0;
}
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", nodes + i);
  sort(nodes, nodes + n);
  if (nodes[n - 1] != n)
    printf("NO\n");
  else
    dodo(0, 0, 0) ? printf("YES\n") : printf("NO\n");
}

```