#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 12;
bool dp[12][14][N][25][3], vis[24];
int a[24], n;
bool bf(int p, int sum, int c) {
  if (p == n) return 1;
  if (sum + 1 == a[p]) {
    if (c != 1) return bf(p + 1, 0, 0);
    return 0;
  }
  int i, x, prev = -1;
  if (sum >= a[p]) return 0;
  for (i = 0; i < p; ++i) {
    x = a[i];
    if (!vis[i] && prev != x) {
      vis[i] = 1;
      if (bf(p, sum + x, c + 1)) return 1;
      vis[i] = 0;
      prev = x;
    }
  }
  return 0;
}
int main() {
  int i;
  cin >> n;
  for (i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  if (bf(0, 0, 0) && a[n - 1] == n) {
    printf("YES\n");
  } else
    printf("NO\n");
}

### Prompt

Please provide a Cpp coded solution 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 N = 1 << 12;
bool dp[12][14][N][25][3], vis[24];
int a[24], n;
bool bf(int p, int sum, int c) {
  if (p == n) return 1;
  if (sum + 1 == a[p]) {
    if (c != 1) return bf(p + 1, 0, 0);
    return 0;
  }
  int i, x, prev = -1;
  if (sum >= a[p]) return 0;
  for (i = 0; i < p; ++i) {
    x = a[i];
    if (!vis[i] && prev != x) {
      vis[i] = 1;
      if (bf(p, sum + x, c + 1)) return 1;
      vis[i] = 0;
      prev = x;
    }
  }
  return 0;
}
int main() {
  int i;
  cin >> n;
  for (i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  if (bf(0, 0, 0) && a[n - 1] == n) {
    printf("YES\n");
  } else
    printf("NO\n");
}

```