#include <bits/stdc++.h>
using namespace std;
int n;
int a[100];
bool vis[100];
bool dfs(int v, int now, int u) {
  if (v == n) return true;
  if (a[v] - 1 == now) {
    if (u != 1)
      return dfs(v + 1, 0, 0);
    else
      return false;
  }
  int la = -1;
  for (int i = 0; i < n; i++)
    if (!vis[i] && a[v] >= a[i] + now && a[i] != la) {
      la = a[i];
      vis[i] = true;
      if (dfs(v, now + a[i], u + 1)) return true;
      vis[i] = false;
    }
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    vis[i] = false;
  }
  sort(a, a + n);
  if (a[n - 1] != n) {
    printf("NO\n");
    return 0;
  }
  if (dfs(0, 0, 0))
    printf("YES\n");
  else
    printf("NO");
  scanf("%d");
}

### Prompt

Generate 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 a[100];
bool vis[100];
bool dfs(int v, int now, int u) {
  if (v == n) return true;
  if (a[v] - 1 == now) {
    if (u != 1)
      return dfs(v + 1, 0, 0);
    else
      return false;
  }
  int la = -1;
  for (int i = 0; i < n; i++)
    if (!vis[i] && a[v] >= a[i] + now && a[i] != la) {
      la = a[i];
      vis[i] = true;
      if (dfs(v, now + a[i], u + 1)) return true;
      vis[i] = false;
    }
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    vis[i] = false;
  }
  sort(a, a + n);
  if (a[n - 1] != n) {
    printf("NO\n");
    return 0;
  }
  if (dfs(0, 0, 0))
    printf("YES\n");
  else
    printf("NO");
  scanf("%d");
}

```