#include <bits/stdc++.h>
using namespace std;
int N, E[30];
bool D[30];
bool dfs(int x, int t) {
  if (E[x] == t) return true;
  for (int i = x - 1; i >= 0; i--)
    if (!D[i] && E[i] <= E[x] - 2 && t + E[i] <= E[x]) {
      D[i] = true;
      if (dfs(x, t + E[i])) return true;
      D[i] = false;
    }
  return false;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &E[i]);
  sort(E, E + N);
  memset(D, false, sizeof(D));
  for (int i = 0; i < N; i++)
    if (!dfs(i, 1)) {
      printf("NO\n");
      return 0;
    }
  for (int i = 0; i < N - 1; i++)
    if (!D[i]) {
      printf("NO\n");
      return 0;
    }
  printf("YES\n");
  return 0;
}

### Prompt

Please create a solution in cpp 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, E[30];
bool D[30];
bool dfs(int x, int t) {
  if (E[x] == t) return true;
  for (int i = x - 1; i >= 0; i--)
    if (!D[i] && E[i] <= E[x] - 2 && t + E[i] <= E[x]) {
      D[i] = true;
      if (dfs(x, t + E[i])) return true;
      D[i] = false;
    }
  return false;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &E[i]);
  sort(E, E + N);
  memset(D, false, sizeof(D));
  for (int i = 0; i < N; i++)
    if (!dfs(i, 1)) {
      printf("NO\n");
      return 0;
    }
  for (int i = 0; i < N - 1; i++)
    if (!D[i]) {
      printf("NO\n");
      return 0;
    }
  printf("YES\n");
  return 0;
}

```