#include <bits/stdc++.h>
using namespace std;
int c[30], n;
void dfs(int u, int pick, int sum, int need, vector<int>& cnt, vector<int>& tot,
         set<vector<int> >& t) {
  if (u > n) {
    if (sum != need) return;
    if (pick < 2) return;
    vector<int> cur = tot;
    for (int i = 0; i < cur.size(); i++) {
      cur[i] -= cnt[i];
    }
    cur[sum + 1]++;
    t.emplace(cur);
    return;
  }
  for (int x = 0; x * u <= need && x <= tot[u]; x++) {
    cnt[u] = x;
    dfs(u + 1, pick + x, sum + x * u, need, cnt, tot, t);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> c[i];
  sort(c, c + n);
  set<vector<int> > s, t;
  s.emplace(vector<int>(n + 1, 0));
  for (int i = 0; i < n; i++) {
    if (c[i] == 1) {
      for (auto x : s) {
        vector<int> arr = x;
        arr[c[i]]++;
        t.emplace(arr);
      }
    } else {
      for (auto x : s) {
        vector<int> arr = x, cnt(n + 1, 0);
        dfs(1, 0, 0, c[i] - 1, cnt, arr, t);
      }
    }
    swap(s, t);
  }
  for (auto x : s) {
    if (accumulate(x.begin(), x.end(), 0) == 1 && x[n] == 1) {
      puts("YES");
      return 0;
    }
  }
  puts("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 c[30], n;
void dfs(int u, int pick, int sum, int need, vector<int>& cnt, vector<int>& tot,
         set<vector<int> >& t) {
  if (u > n) {
    if (sum != need) return;
    if (pick < 2) return;
    vector<int> cur = tot;
    for (int i = 0; i < cur.size(); i++) {
      cur[i] -= cnt[i];
    }
    cur[sum + 1]++;
    t.emplace(cur);
    return;
  }
  for (int x = 0; x * u <= need && x <= tot[u]; x++) {
    cnt[u] = x;
    dfs(u + 1, pick + x, sum + x * u, need, cnt, tot, t);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> c[i];
  sort(c, c + n);
  set<vector<int> > s, t;
  s.emplace(vector<int>(n + 1, 0));
  for (int i = 0; i < n; i++) {
    if (c[i] == 1) {
      for (auto x : s) {
        vector<int> arr = x;
        arr[c[i]]++;
        t.emplace(arr);
      }
    } else {
      for (auto x : s) {
        vector<int> arr = x, cnt(n + 1, 0);
        dfs(1, 0, 0, c[i] - 1, cnt, arr, t);
      }
    }
    swap(s, t);
  }
  for (auto x : s) {
    if (accumulate(x.begin(), x.end(), 0) == 1 && x[n] == 1) {
      puts("YES");
      return 0;
    }
  }
  puts("NO");
  return 0;
}

```