#include <bits/stdc++.h>
using namespace std;
int a[26];
int n;
bool vis[26];
bool dfs(int pre, int val, int child) {
  if (pre == n) return true;
  if (val + 1 == a[pre]) {
    if (child != 1 && dfs(pre + 1, 0, 0)) return true;
    return false;
  }
  int tot = 0;
  for (int i = 0; i < n; i++) {
    if (!vis[i] && val + a[i] < a[pre] && tot != a[i]) {
      tot = a[i];
      vis[i] = true;
      if (dfs(pre, val + a[i], child + 1)) return true;
      vis[i] = false;
    }
  }
  return false;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  memset(vis, false, sizeof(vis));
  if (a[n - 1] == n && dfs(0, 0, 0))
    puts("YES");
  else
    puts("NO");
  return 0;
}

### Prompt

In CPP, your task is to solve 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 a[26];
int n;
bool vis[26];
bool dfs(int pre, int val, int child) {
  if (pre == n) return true;
  if (val + 1 == a[pre]) {
    if (child != 1 && dfs(pre + 1, 0, 0)) return true;
    return false;
  }
  int tot = 0;
  for (int i = 0; i < n; i++) {
    if (!vis[i] && val + a[i] < a[pre] && tot != a[i]) {
      tot = a[i];
      vis[i] = true;
      if (dfs(pre, val + a[i], child + 1)) return true;
      vis[i] = false;
    }
  }
  return false;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  memset(vis, false, sizeof(vis));
  if (a[n - 1] == n && dfs(0, 0, 0))
    puts("YES");
  else
    puts("NO");
  return 0;
}

```