#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
int n, ind, dp[16][25][1 << 16][2], tw[20];
vector<int> v;
int rec(int p, int rem, int mask, bool x) {
  if (p == v.size()) return 0;
  if (~dp[p][rem][mask][x]) return dp[p][rem][mask][x];
  if (rem == 1) {
    if (p + 1 < v.size())
      return dp[p][rem][mask][x] = rec(p + 1, v[p + 1], mask, 0);
    else
      return dp[p][rem][mask][x] = rec(p + 1, 0, mask, 0);
  }
  int ret = 1 + rec(p, rem - 1, mask, 1);
  for (int i = 0; i < v.size(); i++)
    if (!(mask & tw[i])) {
      if ((v[i] < rem - 1) || (x && v[i] <= rem - 1)) {
        ret = min(ret, rec(p, rem - v[i], mask | tw[i], 1));
      }
    }
  return dp[p][rem][mask][x] = ret;
}
int main() {
  for (int i = 0; i < 20; i++) tw[i] = 1 << i;
  cin >> n;
  int t;
  for (int i = 0; i < n; i++) {
    cin >> t;
    v.push_back(t);
  }
  if (n == 1) {
    cout << "YES";
    return 0;
  }
  sort((v).begin(), (v).end());
  for (int i = 0; i < n; i++)
    if (v[i] == 2) {
      cout << "NO";
      return 0;
    }
  int o = 0;
  vector<int> nx;
  for (int i = 0; i < n; i++)
    if (v[i] == 1)
      o++;
    else
      nx.push_back(v[i]);
  if (n - o >= 16 || o == n) {
    cout << "NO";
    return 0;
  }
  v = nx;
  memset((dp), (-1), sizeof(dp));
  int x = rec(0, v[0], 0, 0);
  if (x == o)
    cout << "YES";
  else
    cout << "NO";
  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>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
int n, ind, dp[16][25][1 << 16][2], tw[20];
vector<int> v;
int rec(int p, int rem, int mask, bool x) {
  if (p == v.size()) return 0;
  if (~dp[p][rem][mask][x]) return dp[p][rem][mask][x];
  if (rem == 1) {
    if (p + 1 < v.size())
      return dp[p][rem][mask][x] = rec(p + 1, v[p + 1], mask, 0);
    else
      return dp[p][rem][mask][x] = rec(p + 1, 0, mask, 0);
  }
  int ret = 1 + rec(p, rem - 1, mask, 1);
  for (int i = 0; i < v.size(); i++)
    if (!(mask & tw[i])) {
      if ((v[i] < rem - 1) || (x && v[i] <= rem - 1)) {
        ret = min(ret, rec(p, rem - v[i], mask | tw[i], 1));
      }
    }
  return dp[p][rem][mask][x] = ret;
}
int main() {
  for (int i = 0; i < 20; i++) tw[i] = 1 << i;
  cin >> n;
  int t;
  for (int i = 0; i < n; i++) {
    cin >> t;
    v.push_back(t);
  }
  if (n == 1) {
    cout << "YES";
    return 0;
  }
  sort((v).begin(), (v).end());
  for (int i = 0; i < n; i++)
    if (v[i] == 2) {
      cout << "NO";
      return 0;
    }
  int o = 0;
  vector<int> nx;
  for (int i = 0; i < n; i++)
    if (v[i] == 1)
      o++;
    else
      nx.push_back(v[i]);
  if (n - o >= 16 || o == n) {
    cout << "NO";
    return 0;
  }
  v = nx;
  memset((dp), (-1), sizeof(dp));
  int x = rec(0, v[0], 0, 0);
  if (x == o)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

```