#include <bits/stdc++.h>
using namespace std;
int v[24], aux[24], sons[24], sons1[24];
int n, i, x, cnt, ok, curr, cnt1;
bool bkt(int k) {
  if (v[k] == 1) {
    int it;
    for (it = 0; it < curr; ++it) sons1[it] = sons[it];
    int nr = cnt1;
    for (it = 0; it < curr; ++it) {
      if (aux[it]) {
        if (nr < aux[it]) return false;
        nr -= aux[it];
        sons1[it] += aux[it];
      }
      if (sons1[it] < 2) {
        return false;
      }
    }
    return true;
  }
  bool checker = false;
  for (int it = 0; it < curr; ++it) {
    if (v[k] <= aux[it]) {
      aux[it] -= v[k];
      ++sons[it];
      if (v[k] > 1) {
        aux[curr] = v[k] - 1;
        ++curr;
      }
      checker = bkt(k + 1);
      if (checker) return true;
      if (v[k] > 1) {
        --curr;
        aux[curr] = 0;
      }
      --sons[it];
      aux[it] += v[k];
    }
  }
  return false;
}
int main() {
  cin >> n;
  for (i = 0; i < n; ++i) {
    cin >> x;
    if (x == 2) {
      cout << "NO";
      return 0;
    }
    if (x == n) ++cnt;
    if (x == 1) ++cnt1;
    v[i] = x;
  }
  if (n == 1 && v[0] == 1) {
    cout << "YES";
    return 0;
  }
  if (cnt != 1) {
    cout << "NO";
    return 0;
  }
  sort(v, v + n, greater<int>());
  aux[0] = v[0] - 1;
  curr = 1;
  ok = bkt(1);
  if (ok)
    cout << "YES";
  else
    cout << "NO";
  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;
int v[24], aux[24], sons[24], sons1[24];
int n, i, x, cnt, ok, curr, cnt1;
bool bkt(int k) {
  if (v[k] == 1) {
    int it;
    for (it = 0; it < curr; ++it) sons1[it] = sons[it];
    int nr = cnt1;
    for (it = 0; it < curr; ++it) {
      if (aux[it]) {
        if (nr < aux[it]) return false;
        nr -= aux[it];
        sons1[it] += aux[it];
      }
      if (sons1[it] < 2) {
        return false;
      }
    }
    return true;
  }
  bool checker = false;
  for (int it = 0; it < curr; ++it) {
    if (v[k] <= aux[it]) {
      aux[it] -= v[k];
      ++sons[it];
      if (v[k] > 1) {
        aux[curr] = v[k] - 1;
        ++curr;
      }
      checker = bkt(k + 1);
      if (checker) return true;
      if (v[k] > 1) {
        --curr;
        aux[curr] = 0;
      }
      --sons[it];
      aux[it] += v[k];
    }
  }
  return false;
}
int main() {
  cin >> n;
  for (i = 0; i < n; ++i) {
    cin >> x;
    if (x == 2) {
      cout << "NO";
      return 0;
    }
    if (x == n) ++cnt;
    if (x == 1) ++cnt1;
    v[i] = x;
  }
  if (n == 1 && v[0] == 1) {
    cout << "YES";
    return 0;
  }
  if (cnt != 1) {
    cout << "NO";
    return 0;
  }
  sort(v, v + n, greater<int>());
  aux[0] = v[0] - 1;
  curr = 1;
  ok = bkt(1);
  if (ok)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

```