#include <bits/stdc++.h>
using namespace std;
const int N = 24;
int n, c[N];
bool vis[N];
bool dfs(int u, int v, int cnt, int now) {
  if (u == n) return 1;
  if (c[u] == 1 + now) {
    if (cnt == 1) return 0;
    return dfs(u + 1, 0, 0, 0);
  }
  if (now + c[v] + 1 > c[u]) return 0;
  if (dfs(u, v + 1, cnt, now)) return 1;
  if (vis[v]) return 0;
  vis[v] = 1;
  if (dfs(u, v + 1, cnt + 1, now + c[v])) return 1;
  vis[v] = 0;
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", c + i);
  sort(c, c + n);
  puts(c[n - 1] == n && dfs(0, 0, 0, 0) ? "YES" : "NO");
}

### Prompt

Please formulate 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 N = 24;
int n, c[N];
bool vis[N];
bool dfs(int u, int v, int cnt, int now) {
  if (u == n) return 1;
  if (c[u] == 1 + now) {
    if (cnt == 1) return 0;
    return dfs(u + 1, 0, 0, 0);
  }
  if (now + c[v] + 1 > c[u]) return 0;
  if (dfs(u, v + 1, cnt, now)) return 1;
  if (vis[v]) return 0;
  vis[v] = 1;
  if (dfs(u, v + 1, cnt + 1, now + c[v])) return 1;
  vis[v] = 0;
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", c + i);
  sort(c, c + n);
  puts(c[n - 1] == n && dfs(0, 0, 0, 0) ? "YES" : "NO");
}

```