#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
int c[50];
int cnt[50];
int n;
bool dfs(int k) {
  if (k == 0) {
    for (int i = 1; i <= n; i++) {
      if (cnt[i] != c[i] - 1)
        return false;
      else
        return true;
    }
  }
  for (int i = n; i > k; i--) {
    if (c[i] - c[k] >= 2 && c[k] + cnt[i] < c[i]) {
      cnt[i] += c[k];
      if (dfs(k - 1)) return true;
      cnt[i] -= c[k];
    }
  }
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
  }
  sort(c + 1, c + 1 + n);
  if (n == 1) {
    if (c[n] == 1)
      printf("YES\n");
    else
      printf("NO\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (c[i] == 2) {
      printf("NO\n");
      return 0;
    }
  }
  if (c[n] == n && dfs(n - 1))
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma warning(disable : 4996)
using namespace std;
int c[50];
int cnt[50];
int n;
bool dfs(int k) {
  if (k == 0) {
    for (int i = 1; i <= n; i++) {
      if (cnt[i] != c[i] - 1)
        return false;
      else
        return true;
    }
  }
  for (int i = n; i > k; i--) {
    if (c[i] - c[k] >= 2 && c[k] + cnt[i] < c[i]) {
      cnt[i] += c[k];
      if (dfs(k - 1)) return true;
      cnt[i] -= c[k];
    }
  }
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
  }
  sort(c + 1, c + 1 + n);
  if (n == 1) {
    if (c[n] == 1)
      printf("YES\n");
    else
      printf("NO\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (c[i] == 2) {
      printf("NO\n");
      return 0;
    }
  }
  if (c[n] == n && dfs(n - 1))
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

```