#include <bits/stdc++.h>
using namespace std;
int n;
int a[25];
bool visited[25];
bool dfs(int val, int pos);
int main() {
  int i, j, k;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  memset(visited, false, sizeof(visited));
  for (i = 0; i < n; i++) {
    for (j = i - 1; j >= 0; j--) {
      if (a[i] >= a[j] + 2) break;
    }
    if (!dfs(a[i], j)) {
      cout << "NO" << endl;
      return 0;
    } else {
    }
  }
  for (i = 0; i < n - 1; i++) {
    if (!visited[i]) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  return 0;
}
bool dfs(int val, int pos) {
  if (val == 1) return true;
  int i;
  for (i = pos; i >= 0; i--) {
    if (visited[i]) continue;
    if (val - a[i] < 1) continue;
    visited[i] = true;
    if (dfs(val - a[i], i - 1))
      return true;
    else
      visited[i] = false;
  }
  return false;
}

### 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 n;
int a[25];
bool visited[25];
bool dfs(int val, int pos);
int main() {
  int i, j, k;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  memset(visited, false, sizeof(visited));
  for (i = 0; i < n; i++) {
    for (j = i - 1; j >= 0; j--) {
      if (a[i] >= a[j] + 2) break;
    }
    if (!dfs(a[i], j)) {
      cout << "NO" << endl;
      return 0;
    } else {
    }
  }
  for (i = 0; i < n - 1; i++) {
    if (!visited[i]) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  return 0;
}
bool dfs(int val, int pos) {
  if (val == 1) return true;
  int i;
  for (i = pos; i >= 0; i--) {
    if (visited[i]) continue;
    if (val - a[i] < 1) continue;
    visited[i] = true;
    if (dfs(val - a[i], i - 1))
      return true;
    else
      visited[i] = false;
  }
  return false;
}

```