#include <bits/stdc++.h>
using namespace std;
int c[111];
int n;
vector<int> v[1 << 14];
int dp[1 << 14][14][14];
int rec(int mask, int pos, int leaf) {
  if (pos == n) {
    if (mask == (1 << (n - 1)) and leaf == 0) {
      return 2;
    }
    return 1;
  }
  if (dp[mask][pos][leaf] != 0) {
    return dp[mask][pos][leaf];
  }
  dp[mask][pos][leaf] = 1;
  int want = c[pos];
  int second = 0;
  for (auto submask : v[mask]) {
    second = 0;
    for (int i = 0; i < n; i++) {
      if (submask & (1 << i)) {
        second += c[i];
      }
    }
    if (second <= c[pos] - 1) {
      want = c[pos] - second - 1;
      if (want <= leaf and __builtin_popcount(submask) + want >= 2 and
          rec((mask ^ submask) | (1 << pos), pos + 1, leaf - want) == 2) {
        dp[mask][pos][leaf] = 2;
        return 2;
      }
    }
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  vector<int> second;
  for (int i = 0; i < n; i++) cin >> c[i], second.push_back(c[i]);
  if (n == 1) {
    if (c[0] == 1) {
      cout << "YES";
      return 0;
    }
  }
  sort(second.begin(), second.end());
  int leaf = 0;
  while (second.size() > 0 and second[0] == 1) {
    second.erase(second.begin());
    leaf++;
  }
  if (leaf < (n + 1) / 2) {
    cout << "NO";
    return 0;
  }
  n = second.size();
  if (n == 0) {
    cout << "NO";
    return 0;
  }
  for (int mask = 0; mask < (1 << n); mask++) {
    for (int mask1 = 0; mask1 <= mask; mask1++) {
      if (__builtin_popcount(mask) ==
          __builtin_popcount(mask1) + __builtin_popcount(mask1 ^ mask))
        v[mask].push_back(mask1);
    }
  }
  for (int i = 0; i < n; i++) c[i] = second[i];
  if (rec(0, 0, leaf) == 2) {
    cout << "YES";
  } else {
    cout << "NO";
  }
}

### 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 c[111];
int n;
vector<int> v[1 << 14];
int dp[1 << 14][14][14];
int rec(int mask, int pos, int leaf) {
  if (pos == n) {
    if (mask == (1 << (n - 1)) and leaf == 0) {
      return 2;
    }
    return 1;
  }
  if (dp[mask][pos][leaf] != 0) {
    return dp[mask][pos][leaf];
  }
  dp[mask][pos][leaf] = 1;
  int want = c[pos];
  int second = 0;
  for (auto submask : v[mask]) {
    second = 0;
    for (int i = 0; i < n; i++) {
      if (submask & (1 << i)) {
        second += c[i];
      }
    }
    if (second <= c[pos] - 1) {
      want = c[pos] - second - 1;
      if (want <= leaf and __builtin_popcount(submask) + want >= 2 and
          rec((mask ^ submask) | (1 << pos), pos + 1, leaf - want) == 2) {
        dp[mask][pos][leaf] = 2;
        return 2;
      }
    }
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  vector<int> second;
  for (int i = 0; i < n; i++) cin >> c[i], second.push_back(c[i]);
  if (n == 1) {
    if (c[0] == 1) {
      cout << "YES";
      return 0;
    }
  }
  sort(second.begin(), second.end());
  int leaf = 0;
  while (second.size() > 0 and second[0] == 1) {
    second.erase(second.begin());
    leaf++;
  }
  if (leaf < (n + 1) / 2) {
    cout << "NO";
    return 0;
  }
  n = second.size();
  if (n == 0) {
    cout << "NO";
    return 0;
  }
  for (int mask = 0; mask < (1 << n); mask++) {
    for (int mask1 = 0; mask1 <= mask; mask1++) {
      if (__builtin_popcount(mask) ==
          __builtin_popcount(mask1) + __builtin_popcount(mask1 ^ mask))
        v[mask].push_back(mask1);
    }
  }
  for (int i = 0; i < n; i++) c[i] = second[i];
  if (rec(0, 0, leaf) == 2) {
    cout << "YES";
  } else {
    cout << "NO";
  }
}

```