#include <bits/stdc++.h>
using namespace std;
int n, w[24];
int lg2[1 << 24];
vector<int> v[30];
bitset<1 << 24> f;
vector<int> rv[30];
void DFS(int sum, int mask, int pos) {
  if (sum > n - 1) return;
  if (pos == n + 1) return;
  int nsum = sum;
  int nmask = mask;
  DFS(sum, mask, pos + 1);
  for (auto use : v[pos]) {
    nmask |= 1 << use;
    nsum += pos;
    if (nsum > n) break;
    if ((nmask & -nmask) != nmask) {
      rv[nsum].push_back(nmask);
    }
    DFS(nsum, nmask, pos + 1);
  }
}
void DFS2(int sum, int mask, int use) {
  if (use == n) return;
  DFS2(sum, mask, use + 1);
  sum += w[use];
  mask |= 1 << use;
  if (sum < n) {
    if ((mask & -mask) != mask) rv[sum].push_back(mask);
    DFS2(sum, mask, use + 1);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> w[i];
  sort(w, w + n);
  for (int i = 0; i < n; i++) v[w[i]].push_back(i);
  DFS2(0, 0, 0);
  for (int i = 2; i < (1 << n); i++) lg2[i] = lg2[i / 2] + 1;
  int st = 0;
  while (st < n && w[st] == 1) st++;
  f[(1 << st) - 1] = 1;
  for (int i = (1 << st) - 1; i < (1 << (n - 1)); i++)
    if (f[i]) {
      int cur = lg2[i] + 1;
      for (auto j : rv[w[cur] - 1])
        if ((j & i) == j) f[(i ^ j) | (1 << cur)] = 1;
    }
  cout << (f[1 << (n - 1)] ? "YES" : "NO");
}

### Prompt

Please formulate 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>
using namespace std;
int n, w[24];
int lg2[1 << 24];
vector<int> v[30];
bitset<1 << 24> f;
vector<int> rv[30];
void DFS(int sum, int mask, int pos) {
  if (sum > n - 1) return;
  if (pos == n + 1) return;
  int nsum = sum;
  int nmask = mask;
  DFS(sum, mask, pos + 1);
  for (auto use : v[pos]) {
    nmask |= 1 << use;
    nsum += pos;
    if (nsum > n) break;
    if ((nmask & -nmask) != nmask) {
      rv[nsum].push_back(nmask);
    }
    DFS(nsum, nmask, pos + 1);
  }
}
void DFS2(int sum, int mask, int use) {
  if (use == n) return;
  DFS2(sum, mask, use + 1);
  sum += w[use];
  mask |= 1 << use;
  if (sum < n) {
    if ((mask & -mask) != mask) rv[sum].push_back(mask);
    DFS2(sum, mask, use + 1);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> w[i];
  sort(w, w + n);
  for (int i = 0; i < n; i++) v[w[i]].push_back(i);
  DFS2(0, 0, 0);
  for (int i = 2; i < (1 << n); i++) lg2[i] = lg2[i / 2] + 1;
  int st = 0;
  while (st < n && w[st] == 1) st++;
  f[(1 << st) - 1] = 1;
  for (int i = (1 << st) - 1; i < (1 << (n - 1)); i++)
    if (f[i]) {
      int cur = lg2[i] + 1;
      for (auto j : rv[w[cur] - 1])
        if ((j & i) == j) f[(i ^ j) | (1 << cur)] = 1;
    }
  cout << (f[1 << (n - 1)] ? "YES" : "NO");
}

```