#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const double INF = 1e12, EPS = 1e-9;
int n, out;
vector<int> in;
int dp[12][1 << 11];
bool check() {
  for (int i = 0; i < (int)n + 1; i++)
    for (int j = 0; j < (int)1 << n; j++) dp[i][j] = inf;
  for (int i = 0; i < (int)n; i++)
    if (in[i] >= 3) dp[1][1 << i] = in[i];
  for (int i = 0; i < (int)1 << n; i++) {
    for (int j = 0; j < (int)n; j++)
      for (int k = (i - 1) & i; k; k = (k - 1) & i) {
        if (dp[j][k] < inf && dp[1][i ^ k] < inf)
          dp[j + 1][i] = min(dp[j + 1][i], dp[j][k] + dp[1][i ^ k]);
      }
    for (int j = 0; j < (int)n; j++)
      if (!(i & 1 << j))
        for (int k = 0; k < (int)n; k++) {
          if (in[j] - 1 < dp[k][i]) continue;
          if (in[j] - 1 - dp[k][i] + k < 2) continue;
          dp[1][i | 1 << j] = in[j];
        }
  }
  return dp[1][(1 << n) - 1] < inf && in[n - 1] == n + out;
}
int main() {
  cin >> n;
  for (int i = 0; i < (int)n; i++) {
    int x;
    cin >> x;
    if (x == 1)
      out++;
    else
      in.push_back(x);
  }
  if (n == out || in.size() > 11) {
    cout << (n == 1 ? "YES" : "NO") << endl;
    return 0;
  }
  sort((in).begin(), (in).end());
  n = in.size();
  cout << (check() ? "YES" : "NO") << endl;
  return 0;
}

### 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 inf = (int)1e9;
const double INF = 1e12, EPS = 1e-9;
int n, out;
vector<int> in;
int dp[12][1 << 11];
bool check() {
  for (int i = 0; i < (int)n + 1; i++)
    for (int j = 0; j < (int)1 << n; j++) dp[i][j] = inf;
  for (int i = 0; i < (int)n; i++)
    if (in[i] >= 3) dp[1][1 << i] = in[i];
  for (int i = 0; i < (int)1 << n; i++) {
    for (int j = 0; j < (int)n; j++)
      for (int k = (i - 1) & i; k; k = (k - 1) & i) {
        if (dp[j][k] < inf && dp[1][i ^ k] < inf)
          dp[j + 1][i] = min(dp[j + 1][i], dp[j][k] + dp[1][i ^ k]);
      }
    for (int j = 0; j < (int)n; j++)
      if (!(i & 1 << j))
        for (int k = 0; k < (int)n; k++) {
          if (in[j] - 1 < dp[k][i]) continue;
          if (in[j] - 1 - dp[k][i] + k < 2) continue;
          dp[1][i | 1 << j] = in[j];
        }
  }
  return dp[1][(1 << n) - 1] < inf && in[n - 1] == n + out;
}
int main() {
  cin >> n;
  for (int i = 0; i < (int)n; i++) {
    int x;
    cin >> x;
    if (x == 1)
      out++;
    else
      in.push_back(x);
  }
  if (n == out || in.size() > 11) {
    cout << (n == 1 ? "YES" : "NO") << endl;
    return 0;
  }
  sort((in).begin(), (in).end());
  n = in.size();
  cout << (check() ? "YES" : "NO") << endl;
  return 0;
}

```