#include <bits/stdc++.h>
using namespace std;
int N, cnt;
vector<int> C, P, Q;
vector<vector<int> > cc;
int dp(int mask, int c) {
  int &ret = cc[mask][c];
  if (ret != -1) return ret;
  if (mask & (1 << (N - 1))) return ret = ((mask == (1 << (N - 1))) & (c == 0));
  int idx = 0;
  for (int i = N - 1; i >= 0; i--) {
    if (mask & (1 << i)) {
      idx = i + 1;
      break;
    }
  }
  for (int i = 0; i <= c; i++) {
    for (int j = 0; j < (1 << N); j++) {
      if ((mask | j) == mask && P[j] + 1 + i == C[idx] && Q[j] + i >= 2) {
        if (dp(mask ^ j ^ (1 << idx), c - i)) return ret = 1;
      }
    }
  }
  return ret = 0;
}
int main() {
  scanf("%d", &N);
  cnt = 0;
  for (int i = 0; i < N; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1)
      cnt++;
    else
      C.push_back(t);
  }
  sort(C.begin(), C.end());
  N = C.size();
  if (N == 0) {
    if (cnt == 1)
      printf("YES");
    else
      printf("NO");
    return 0;
  }
  if (N > 12) {
    printf("NO");
    return 0;
  }
  P.resize((1 << N));
  Q.resize((1 << N));
  for (int i = 0; i < (1 << N); i++) {
    P[i] = 0;
    Q[i] = 0;
    for (int j = 0; j < N; j++) {
      if (i & (1 << j)) {
        P[i] += C[j];
        Q[i] += 1;
      }
    }
  }
  cc = vector<vector<int> >((1 << N), vector<int>(cnt + 1, -1));
  if (dp(0, cnt))
    printf("YES");
  else
    printf("NO");
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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, cnt;
vector<int> C, P, Q;
vector<vector<int> > cc;
int dp(int mask, int c) {
  int &ret = cc[mask][c];
  if (ret != -1) return ret;
  if (mask & (1 << (N - 1))) return ret = ((mask == (1 << (N - 1))) & (c == 0));
  int idx = 0;
  for (int i = N - 1; i >= 0; i--) {
    if (mask & (1 << i)) {
      idx = i + 1;
      break;
    }
  }
  for (int i = 0; i <= c; i++) {
    for (int j = 0; j < (1 << N); j++) {
      if ((mask | j) == mask && P[j] + 1 + i == C[idx] && Q[j] + i >= 2) {
        if (dp(mask ^ j ^ (1 << idx), c - i)) return ret = 1;
      }
    }
  }
  return ret = 0;
}
int main() {
  scanf("%d", &N);
  cnt = 0;
  for (int i = 0; i < N; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1)
      cnt++;
    else
      C.push_back(t);
  }
  sort(C.begin(), C.end());
  N = C.size();
  if (N == 0) {
    if (cnt == 1)
      printf("YES");
    else
      printf("NO");
    return 0;
  }
  if (N > 12) {
    printf("NO");
    return 0;
  }
  P.resize((1 << N));
  Q.resize((1 << N));
  for (int i = 0; i < (1 << N); i++) {
    P[i] = 0;
    Q[i] = 0;
    for (int j = 0; j < N; j++) {
      if (i & (1 << j)) {
        P[i] += C[j];
        Q[i] += 1;
      }
    }
  }
  cc = vector<vector<int> >((1 << N), vector<int>(cnt + 1, -1));
  if (dp(0, cnt))
    printf("YES");
  else
    printf("NO");
}

```