#include <bits/stdc++.h>
using namespace std;
const int maxn = 30;
int n, c[maxn], anss = 0;
char ans[2][5] = {"NO", "YES"};
int rem[maxn], s[maxn];
bool dfs(int cur) {
  int i;
  if (!cur || c[cur] == 1) {
    for (i = cur + 1; i <= n; i++)
      if (rem[i] == 1 && !s[i]) return 0;
    return 1;
  }
  rem[cur] = c[cur] - 1;
  s[cur] = 0;
  if (cur == n) return c[n] != n ? 0 : dfs(n - 1);
  for (i = n; i > cur; i--)
    if (rem[i] >= c[cur] + (!s[i]) * c[1]) {
      s[i]++;
      rem[i] -= c[cur];
      if (dfs(cur - 1)) return 1;
      rem[i] += c[cur];
      s[i]--;
    }
  return 0;
}
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &c[i]);
  sort(c + 1, c + 1 + n);
  anss = dfs(n);
  printf("%s\n", ans[anss]);
}

### 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>
using namespace std;
const int maxn = 30;
int n, c[maxn], anss = 0;
char ans[2][5] = {"NO", "YES"};
int rem[maxn], s[maxn];
bool dfs(int cur) {
  int i;
  if (!cur || c[cur] == 1) {
    for (i = cur + 1; i <= n; i++)
      if (rem[i] == 1 && !s[i]) return 0;
    return 1;
  }
  rem[cur] = c[cur] - 1;
  s[cur] = 0;
  if (cur == n) return c[n] != n ? 0 : dfs(n - 1);
  for (i = n; i > cur; i--)
    if (rem[i] >= c[cur] + (!s[i]) * c[1]) {
      s[i]++;
      rem[i] -= c[cur];
      if (dfs(cur - 1)) return 1;
      rem[i] += c[cur];
      s[i]--;
    }
  return 0;
}
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &c[i]);
  sort(c + 1, c + 1 + n);
  anss = dfs(n);
  printf("%s\n", ans[anss]);
}

```