#include <bits/stdc++.h>
using namespace std;
int a[26];
bool v[26], flag;
int n;
int main() {
  int i, j, s;
  scanf("%d", &n);
  for (i = 0; i < n; ++i) scanf("%d", &a[i]);
  sort(a, a + n);
  for (i = 0; i < n; ++i)
    for (s = 1, j = i - 1; j >= 0 && s < a[i]; --j)
      if (!v[j] && a[j] < a[i] - 1 && s + a[j] <= a[i]) {
        v[j] = 1;
        s += a[j];
      }
  for (i = 0; i < n - 1; ++i)
    if (!v[i]) {
      flag = 1;
      break;
    }
  puts(flag ? "NO" : "YES");
  return 0;
}

### Prompt

Create a solution in Cpp for 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];
bool v[26], flag;
int n;
int main() {
  int i, j, s;
  scanf("%d", &n);
  for (i = 0; i < n; ++i) scanf("%d", &a[i]);
  sort(a, a + n);
  for (i = 0; i < n; ++i)
    for (s = 1, j = i - 1; j >= 0 && s < a[i]; --j)
      if (!v[j] && a[j] < a[i] - 1 && s + a[j] <= a[i]) {
        v[j] = 1;
        s += a[j];
      }
  for (i = 0; i < n - 1; ++i)
    if (!v[i]) {
      flag = 1;
      break;
    }
  puts(flag ? "NO" : "YES");
  return 0;
}

```