#include <bits/stdc++.h>
using namespace std;
int nxtNode[25], has[25], need[25];
int n;
bool bt(int u, int nodes) {
  if (u == n + 1) {
    for (int i = 1; i <= n; ++i)
      if (has[i] != need[i]) return 0;
    return 1;
  }
  while (nxtNode[nodes] == 0) --nodes;
  has[u] = 1;
  need[u] = nodes;
  bool ans = 0;
  for (int v = 1; v < u; ++v) {
    if (has[v] + nodes <= need[v] && nodes + 1 < need[v]) {
      has[v] += nodes;
      --nxtNode[nodes];
      ans = bt(u + 1, nodes) || ans;
      has[v] -= nodes;
      ++nxtNode[nodes];
      if (nodes == 1) break;
    }
  }
  return ans;
}
int main() {
  memset(nxtNode, 0, sizeof nxtNode);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    ++nxtNode[x];
  }
  has[1] = 1;
  need[1] = n;
  if (bt(2, n - 1))
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

### Prompt

Develop a solution in CPP 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;
int nxtNode[25], has[25], need[25];
int n;
bool bt(int u, int nodes) {
  if (u == n + 1) {
    for (int i = 1; i <= n; ++i)
      if (has[i] != need[i]) return 0;
    return 1;
  }
  while (nxtNode[nodes] == 0) --nodes;
  has[u] = 1;
  need[u] = nodes;
  bool ans = 0;
  for (int v = 1; v < u; ++v) {
    if (has[v] + nodes <= need[v] && nodes + 1 < need[v]) {
      has[v] += nodes;
      --nxtNode[nodes];
      ans = bt(u + 1, nodes) || ans;
      has[v] -= nodes;
      ++nxtNode[nodes];
      if (nodes == 1) break;
    }
  }
  return ans;
}
int main() {
  memset(nxtNode, 0, sizeof nxtNode);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    ++nxtNode[x];
  }
  has[1] = 1;
  need[1] = n;
  if (bt(2, n - 1))
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

```