#include <bits/stdc++.h>
using namespace std;
int n, a[30], v[30];
bool cal(int p, int sum, int child) {
  if (p == n) {
    return true;
  }
  if (sum + 1 == a[p]) {
    if (child != 1)
      return cal(p + 1, 0, 0);
    else
      return false;
  }
  int l = -1;
  for (int x = 0; x < n; x++) {
    if (v[x] == 0 && a[x] + sum < a[p] && l != a[x]) {
      l = a[x];
      v[x] = 1;
      if (cal(p, sum + a[x], child + 1)) return true;
      v[x] = 0;
    }
  }
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  sort(a, a + n);
  if (a[n - 1] == n && cal(0, 0, 0))
    printf("YES\n");
  else
    printf("NO\n");
}

### 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, a[30], v[30];
bool cal(int p, int sum, int child) {
  if (p == n) {
    return true;
  }
  if (sum + 1 == a[p]) {
    if (child != 1)
      return cal(p + 1, 0, 0);
    else
      return false;
  }
  int l = -1;
  for (int x = 0; x < n; x++) {
    if (v[x] == 0 && a[x] + sum < a[p] && l != a[x]) {
      l = a[x];
      v[x] = 1;
      if (cal(p, sum + a[x], child + 1)) return true;
      v[x] = 0;
    }
  }
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  sort(a, a + n);
  if (a[n - 1] == n && cal(0, 0, 0))
    printf("YES\n");
  else
    printf("NO\n");
}

```