#include <bits/stdc++.h>
using namespace std;
short int cnt[1 << 24];
short int bits[1 << 24];
const int N = 24;
int n, ara[N];
map<int, int> dp[25];
int call(int pos, int mask) {
  if (mask == (1 << (n - 1)) - 1) {
    return ara[n - 1] == n;
  }
  if (pos == n) return 0;
  if (dp[pos].find(mask) != dp[pos].end()) return dp[pos][mask];
  int ret = 0;
  int my = 0;
  for (int i = 0; i < pos; i++) {
    if (mask & (1 << i)) continue;
    my = my | (1 << i);
  }
  for (int now = my; now > 0; now = (now - 1) & my) {
    if (bits[now] < 2) continue;
    if (cnt[now] + 1 != ara[pos]) continue;
    ret = ret | call(pos + 1, mask | now);
    if (ret == 1) return dp[pos][mask] = 1;
  }
  return dp[pos][mask] = ret;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> ara[i];
  sort(ara, ara + n);
  for (int mask = 0; mask < (1 << n); mask++) {
    for (int j = 0; j < n; j++) {
      if (mask & (1 << j)) cnt[mask] += ara[j];
    }
    bits[mask] = __builtin_popcount(mask);
  }
  int pos = 0;
  for (int i = 0; i < n; i++) {
    if (ara[i] > 1) {
      pos = i;
      break;
    }
  }
  if (call(pos, 0))
    cout << "YES\n";
  else
    cout << "NO\n";
}

### Prompt

Please create a solution in cpp 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>
using namespace std;
short int cnt[1 << 24];
short int bits[1 << 24];
const int N = 24;
int n, ara[N];
map<int, int> dp[25];
int call(int pos, int mask) {
  if (mask == (1 << (n - 1)) - 1) {
    return ara[n - 1] == n;
  }
  if (pos == n) return 0;
  if (dp[pos].find(mask) != dp[pos].end()) return dp[pos][mask];
  int ret = 0;
  int my = 0;
  for (int i = 0; i < pos; i++) {
    if (mask & (1 << i)) continue;
    my = my | (1 << i);
  }
  for (int now = my; now > 0; now = (now - 1) & my) {
    if (bits[now] < 2) continue;
    if (cnt[now] + 1 != ara[pos]) continue;
    ret = ret | call(pos + 1, mask | now);
    if (ret == 1) return dp[pos][mask] = 1;
  }
  return dp[pos][mask] = ret;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> ara[i];
  sort(ara, ara + n);
  for (int mask = 0; mask < (1 << n); mask++) {
    for (int j = 0; j < n; j++) {
      if (mask & (1 << j)) cnt[mask] += ara[j];
    }
    bits[mask] = __builtin_popcount(mask);
  }
  int pos = 0;
  for (int i = 0; i < n; i++) {
    if (ara[i] > 1) {
      pos = i;
      break;
    }
  }
  if (call(pos, 0))
    cout << "YES\n";
  else
    cout << "NO\n";
}

```