#include <bits/stdc++.h>
using namespace std;
const int N = 42;
int c[N], n, vis[N];
int dfs(int idx) {
  vis[idx] = 1;
  if (c[idx] == 1) return 1;
  int cnt = c[idx] - 1;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      if (c[i] <= cnt && c[i] < c[idx] - 1) {
        cnt -= c[i];
        if (!dfs(i)) return 0;
      }
    }
  }
  return (cnt == 0);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", c + i);
  }
  sort(c + 1, c + n + 1);
  reverse(c + 1, c + n + 1);
  int ans = dfs(1);
  if (ans) {
    int s = 0;
    for (int i = 1; i <= n; i++) s += vis[i];
    if (s == n)
      cout << "YES\n";
    else
      cout << "NO\n";
  } else
    cout << "NO\n";
  return 0;
}

### 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 N = 42;
int c[N], n, vis[N];
int dfs(int idx) {
  vis[idx] = 1;
  if (c[idx] == 1) return 1;
  int cnt = c[idx] - 1;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      if (c[i] <= cnt && c[i] < c[idx] - 1) {
        cnt -= c[i];
        if (!dfs(i)) return 0;
      }
    }
  }
  return (cnt == 0);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", c + i);
  }
  sort(c + 1, c + n + 1);
  reverse(c + 1, c + n + 1);
  int ans = dfs(1);
  if (ans) {
    int s = 0;
    for (int i = 1; i <= n; i++) s += vis[i];
    if (s == n)
      cout << "YES\n";
    else
      cout << "NO\n";
  } else
    cout << "NO\n";
  return 0;
}

```