#include <bits/stdc++.h>
using namespace std;
int c[24], vis[24], n;
bool dfs(int p, int child, int number_of_children, int sb_tree) {
  if (p == n) return 1;
  if (c[p] == 1 + sb_tree) {
    if (number_of_children == 1) return 0;
    vis[p] = 0;
    return dfs(p + 1, 0, 0, 0);
  }
  if (child >= p) return 0;
  if (!vis[child]) {
    vis[child] = 1;
    if (c[child] + sb_tree + 1 <= c[p] &&
        dfs(p, child + 1, number_of_children + 1, sb_tree + c[child]))
      return 1;
    vis[child] = 0;
  }
  if (dfs(p, child + 1, number_of_children, sb_tree)) return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", c + i);
  sort(c, c + n);
  if (c[n - 1] == n && dfs(0, 0, 0, 0))
    puts("YES");
  else
    puts("NO");
}

### 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 c[24], vis[24], n;
bool dfs(int p, int child, int number_of_children, int sb_tree) {
  if (p == n) return 1;
  if (c[p] == 1 + sb_tree) {
    if (number_of_children == 1) return 0;
    vis[p] = 0;
    return dfs(p + 1, 0, 0, 0);
  }
  if (child >= p) return 0;
  if (!vis[child]) {
    vis[child] = 1;
    if (c[child] + sb_tree + 1 <= c[p] &&
        dfs(p, child + 1, number_of_children + 1, sb_tree + c[child]))
      return 1;
    vis[child] = 0;
  }
  if (dfs(p, child + 1, number_of_children, sb_tree)) return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", c + i);
  sort(c, c + n);
  if (c[n - 1] == n && dfs(0, 0, 0, 0))
    puts("YES");
  else
    puts("NO");
}

```