#include <bits/stdc++.h>
using namespace std;
int a[30];
bool mark[30];
bool dp[30][1000];
int n;
void update() {
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 1000; j++) {
      dp[i][j] = dp[i - 1][j];
      if (j >= a[i] && !mark[i]) dp[i][j] |= dp[i - 1][j - a[i]];
    }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  update();
  int k = n;
  for (int i = 1; i <= n; i++) {
    int t = a[i] - 1;
    for (int j = i - 1; j >= 1; j--) {
      if (a[j] != a[i] - 1 && a[j] <= t && !mark[j] && dp[j - 1][t - a[j]]) {
        update();
        mark[j] = 1;
        t -= a[j];
        k--;
      }
    }
    if (t) k = 200;
  }
  if (k == 1)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int a[30];
bool mark[30];
bool dp[30][1000];
int n;
void update() {
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 1000; j++) {
      dp[i][j] = dp[i - 1][j];
      if (j >= a[i] && !mark[i]) dp[i][j] |= dp[i - 1][j - a[i]];
    }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  update();
  int k = n;
  for (int i = 1; i <= n; i++) {
    int t = a[i] - 1;
    for (int j = i - 1; j >= 1; j--) {
      if (a[j] != a[i] - 1 && a[j] <= t && !mark[j] && dp[j - 1][t - a[j]]) {
        update();
        mark[j] = 1;
        t -= a[j];
        k--;
      }
    }
    if (t) k = 200;
  }
  if (k == 1)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

```