#include <bits/stdc++.h>
using namespace std;
const int inf = 1 << 28;
const int mod = (int)1e9 + 7;
const double eps = 1e-8;
int n, m, k;
int a[33];
int sum[33];
bool dfs(int x) {
  if (x == n + 1) return true;
  for (int i = 1; i < x; i++) {
    if (sum[i] + a[x] < a[i] && a[x] + 1 != a[i]) {
      sum[i] += a[x];
      if (dfs(x + 1)) return true;
      sum[i] -= a[x];
    }
  }
  return false;
}
int main() {
  while (cin >> n) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    sort(a + 1, a + n + 1, greater<int>());
    if (a[1] != n || count(a + 1, a + n + 1, 2)) return puts("NO");
    if (n == 1 || dfs(2))
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int inf = 1 << 28;
const int mod = (int)1e9 + 7;
const double eps = 1e-8;
int n, m, k;
int a[33];
int sum[33];
bool dfs(int x) {
  if (x == n + 1) return true;
  for (int i = 1; i < x; i++) {
    if (sum[i] + a[x] < a[i] && a[x] + 1 != a[i]) {
      sum[i] += a[x];
      if (dfs(x + 1)) return true;
      sum[i] -= a[x];
    }
  }
  return false;
}
int main() {
  while (cin >> n) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    sort(a + 1, a + n + 1, greater<int>());
    if (a[1] != n || count(a + 1, a + n + 1, 2)) return puts("NO");
    if (n == 1 || dfs(2))
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}

```