#include <bits/stdc++.h>
using namespace std;
int c[25];
bool vis[25];
int n, m;
bool dfs(int p, int s, int k) {
  if (p > n) return m == n - 1;
  if (!k) return dfs(p + 1, p, c[p + 1]);
  for (; s; s--)
    if (!vis[s] && c[s] + 1 < c[p] && c[s] + 1 <= k) {
      vis[s] = true;
      k -= c[s] + 1;
      m++;
      if (dfs(p, s - 1, k)) return true;
      vis[s] = false;
      k += c[s] + 1;
      m--;
    }
  return false;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &c[i]);
      c[i]--;
    }
    sort(c + 1, c + 1 + n);
    memset(vis, false, sizeof(vis));
    m = 0;
    puts(dfs(1, 0, c[1]) ? "YES" : "NO");
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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[25];
bool vis[25];
int n, m;
bool dfs(int p, int s, int k) {
  if (p > n) return m == n - 1;
  if (!k) return dfs(p + 1, p, c[p + 1]);
  for (; s; s--)
    if (!vis[s] && c[s] + 1 < c[p] && c[s] + 1 <= k) {
      vis[s] = true;
      k -= c[s] + 1;
      m++;
      if (dfs(p, s - 1, k)) return true;
      vis[s] = false;
      k += c[s] + 1;
      m--;
    }
  return false;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &c[i]);
      c[i]--;
    }
    sort(c + 1, c + 1 + n);
    memset(vis, false, sizeof(vis));
    m = 0;
    puts(dfs(1, 0, c[1]) ? "YES" : "NO");
  }
  return 0;
}

```