#include <bits/stdc++.h>
int n, c[25], s[25];
bool check(int pos) {
  int cnt = 0;
  for (int i = 0; i < pos; ++i) {
    cnt += c[i];
    if (s[i] == 0 && c[i] == 1) {
      return false;
    }
    if (s[i] == 1 && c[i] == 0) {
      return false;
    }
  }
  return cnt == (n - pos);
}
bool go(int pos) {
  if (c[pos] == 0) {
    return check(pos);
  }
  for (int i = 0; i < pos; ++i) {
    if (s[i] < 2 && c[pos] <= c[i]) {
      if (c[pos] + 1 == c[i] && s[i] == 0) {
        continue;
      }
      s[i] += 1;
      c[i] = c[i] - c[pos] - 1;
      if (go(pos + 1)) {
        return true;
      }
      c[i] = c[i] + c[pos] + 1;
      s[i] -= 1;
    }
  }
  return false;
}
void solve() {
  std::cin >> n;
  for (int i = 0; i < n; ++i) {
    std::cin >> c[i];
    --c[i];
  }
  std::sort(c, c + n);
  std::reverse(c, c + n);
  if (c[0] != n - 1) {
    std::cout << "NO";
    return;
  }
  int cnt_one = 0;
  for (int i = 0; i < n; ++i) {
    if (c[i] == 0) {
      ++cnt_one;
    }
  }
  if (cnt_one < n / 2) {
    std::cout << "NO";
    return;
  }
  if (go(1)) {
    std::cout << "YES";
  } else {
    std::cout << "NO";
  }
  std::cout.flush();
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  solve();
  return 0;
}

### Prompt

Generate a CPP solution to 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>
int n, c[25], s[25];
bool check(int pos) {
  int cnt = 0;
  for (int i = 0; i < pos; ++i) {
    cnt += c[i];
    if (s[i] == 0 && c[i] == 1) {
      return false;
    }
    if (s[i] == 1 && c[i] == 0) {
      return false;
    }
  }
  return cnt == (n - pos);
}
bool go(int pos) {
  if (c[pos] == 0) {
    return check(pos);
  }
  for (int i = 0; i < pos; ++i) {
    if (s[i] < 2 && c[pos] <= c[i]) {
      if (c[pos] + 1 == c[i] && s[i] == 0) {
        continue;
      }
      s[i] += 1;
      c[i] = c[i] - c[pos] - 1;
      if (go(pos + 1)) {
        return true;
      }
      c[i] = c[i] + c[pos] + 1;
      s[i] -= 1;
    }
  }
  return false;
}
void solve() {
  std::cin >> n;
  for (int i = 0; i < n; ++i) {
    std::cin >> c[i];
    --c[i];
  }
  std::sort(c, c + n);
  std::reverse(c, c + n);
  if (c[0] != n - 1) {
    std::cout << "NO";
    return;
  }
  int cnt_one = 0;
  for (int i = 0; i < n; ++i) {
    if (c[i] == 0) {
      ++cnt_one;
    }
  }
  if (cnt_one < n / 2) {
    std::cout << "NO";
    return;
  }
  if (go(1)) {
    std::cout << "YES";
  } else {
    std::cout << "NO";
  }
  std::cout.flush();
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  solve();
  return 0;
}

```