#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> c(n);
  int roots = 0, leaves = 0;
  vector<int> other;
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
    if (c[i] == n) {
      ++roots;
    } else if (c[i] == 1) {
      ++leaves;
    } else {
      if (c[i] == 2) {
        cout << "NO\n";
        return 0;
      }
      other.emplace_back(c[i]);
    }
  }
  if (n == 1) {
    cout << (roots == 1 ? "YES\n" : "NO\n");
    return 0;
  }
  if (roots != 1 || 2 * leaves <= n) {
    cout << "NO\n";
    return 0;
  }
  if (n <= 4) {
    cout << "YES\n";
    return 0;
  }
  sort(other.begin(), other.end());
  vector<int> sums(1 << other.size());
  for (int i = 0; i < (1 << other.size()); ++i) {
    int sum = 0;
    for (int j = 0; j < other.size(); ++j) {
      if (i & (1 << j)) {
        sum += other[j];
      }
    }
    sums[i] = sum;
  }
  other.emplace_back(n);
  int final_mask = 1 << (other.size() - 1);
  --final_mask;
  function<void(int, int, int)> rec = [&](int idx, int left, int used) {
    if (idx == other.size()) {
      if (left == 0 && used == final_mask) {
        cout << "YES\n";
        exit(0);
      }
      return;
    }
    int v = other[idx] - 1;
    int mask = ~used & ((1 << idx) - 1);
    for (int i = mask;; i = (i - 1) & mask) {
      if ((sums[i] == v && __builtin_popcount(i) == 1) || (sums[i] > v) ||
          (v - sums[i] > left)) {
        if (!i) {
          break;
        }
        continue;
      }
      rec(idx + 1, left - (v - sums[i]), used | i);
      if (!i) {
        break;
      }
    }
  };
  rec(0, leaves, 0);
  cout << "NO\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
using ll = long long;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> c(n);
  int roots = 0, leaves = 0;
  vector<int> other;
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
    if (c[i] == n) {
      ++roots;
    } else if (c[i] == 1) {
      ++leaves;
    } else {
      if (c[i] == 2) {
        cout << "NO\n";
        return 0;
      }
      other.emplace_back(c[i]);
    }
  }
  if (n == 1) {
    cout << (roots == 1 ? "YES\n" : "NO\n");
    return 0;
  }
  if (roots != 1 || 2 * leaves <= n) {
    cout << "NO\n";
    return 0;
  }
  if (n <= 4) {
    cout << "YES\n";
    return 0;
  }
  sort(other.begin(), other.end());
  vector<int> sums(1 << other.size());
  for (int i = 0; i < (1 << other.size()); ++i) {
    int sum = 0;
    for (int j = 0; j < other.size(); ++j) {
      if (i & (1 << j)) {
        sum += other[j];
      }
    }
    sums[i] = sum;
  }
  other.emplace_back(n);
  int final_mask = 1 << (other.size() - 1);
  --final_mask;
  function<void(int, int, int)> rec = [&](int idx, int left, int used) {
    if (idx == other.size()) {
      if (left == 0 && used == final_mask) {
        cout << "YES\n";
        exit(0);
      }
      return;
    }
    int v = other[idx] - 1;
    int mask = ~used & ((1 << idx) - 1);
    for (int i = mask;; i = (i - 1) & mask) {
      if ((sums[i] == v && __builtin_popcount(i) == 1) || (sums[i] > v) ||
          (v - sums[i] > left)) {
        if (!i) {
          break;
        }
        continue;
      }
      rec(idx + 1, left - (v - sums[i]), used | i);
      if (!i) {
        break;
      }
    }
  };
  rec(0, leaves, 0);
  cout << "NO\n";
  return 0;
}

```