#include <bits/stdc++.h>
using namespace std;
long n, q;
vector<long> v, v1;
long ans;
vector<long> v0;
void solve(vector<long> v0, vector<long> v1, long ps) {
  if (v0.size() + v1.size() > ps + 2) return;
  sort(v0.begin(), v0.end());
  sort(v1.begin(), v1.end());
  if (ans) return;
  if (ps == -1) {
    if (v0.size() + v1.size() == 0) ans = 1;
    return;
  }
  vector<long> vn0, vn1;
  for (int i = 0; i < v0.size(); i++)
    if (v0[i] > v[ps] && (i == 0 || v0[i] != v0[i - 1])) {
      vn0 = v0;
      vn1 = v1;
      vn0.erase(vn0.begin() + i);
      vn1.push_back(v0[i] - v[ps]);
      if (v[ps] > 1) vn0.push_back(v[ps] - 1);
      solve(vn0, vn1, ps - 1);
    }
  for (int i = 0; i < v1.size(); i++)
    if (v1[i] >= v[ps] && (i == 0 || v1[i] != v1[i - 1])) {
      vn0 = v0;
      vn1 = v1;
      vn1.erase(vn1.begin() + i);
      if (v1[i] > v[ps]) vn1.push_back(v1[i] - v[ps]);
      if (v[ps] > 1) vn0.push_back(v[ps] - 1);
      solve(vn0, vn1, ps - 1);
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> q;
    v.push_back(q);
  }
  sort(v.begin(), v.end());
  long calc = 0;
  for (int i = 0; i < v.size(); i++)
    if (v[i] == 1) ++calc;
  if (calc * 2 >= n - 4) {
    if (v[n - 1] > 1) v0.push_back(v[n - 1] - 1);
    v.pop_back();
    solve(v0, v1, n - 2);
  }
  if (ans)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  cin.get();
  cin.get();
  return 0;
}

### Prompt

Create a solution in cpp for 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;
long n, q;
vector<long> v, v1;
long ans;
vector<long> v0;
void solve(vector<long> v0, vector<long> v1, long ps) {
  if (v0.size() + v1.size() > ps + 2) return;
  sort(v0.begin(), v0.end());
  sort(v1.begin(), v1.end());
  if (ans) return;
  if (ps == -1) {
    if (v0.size() + v1.size() == 0) ans = 1;
    return;
  }
  vector<long> vn0, vn1;
  for (int i = 0; i < v0.size(); i++)
    if (v0[i] > v[ps] && (i == 0 || v0[i] != v0[i - 1])) {
      vn0 = v0;
      vn1 = v1;
      vn0.erase(vn0.begin() + i);
      vn1.push_back(v0[i] - v[ps]);
      if (v[ps] > 1) vn0.push_back(v[ps] - 1);
      solve(vn0, vn1, ps - 1);
    }
  for (int i = 0; i < v1.size(); i++)
    if (v1[i] >= v[ps] && (i == 0 || v1[i] != v1[i - 1])) {
      vn0 = v0;
      vn1 = v1;
      vn1.erase(vn1.begin() + i);
      if (v1[i] > v[ps]) vn1.push_back(v1[i] - v[ps]);
      if (v[ps] > 1) vn0.push_back(v[ps] - 1);
      solve(vn0, vn1, ps - 1);
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> q;
    v.push_back(q);
  }
  sort(v.begin(), v.end());
  long calc = 0;
  for (int i = 0; i < v.size(); i++)
    if (v[i] == 1) ++calc;
  if (calc * 2 >= n - 4) {
    if (v[n - 1] > 1) v0.push_back(v[n - 1] - 1);
    v.pop_back();
    solve(v0, v1, n - 2);
  }
  if (ans)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  cin.get();
  cin.get();
  return 0;
}

```