#include <bits/stdc++.h>
using namespace std;
int N;
int A[50];
int memo[1 << 12][30];
bool solve(int m, int ones) {
  if (m == 0) return ones == 1;
  if (memo[m][ones] != -1) return memo[m][ones] == 1;
  int rt = -1;
  int NN = __builtin_popcount(m) + ones;
  for (int i = 0; i < N; i++) {
    if (m & 1 << i) {
      if (A[i] == NN) {
        rt = i;
      }
    }
  }
  if (rt == -1) return false;
  m ^= 1 << rt;
  NN -= ones + 1;
  vector<vector<bool> > DP(1 << NN, vector<bool>(30, false));
  DP[0][0] = true;
  for (int j = 0, ss = m; j < 1 << NN; j++, ss = ss - 1 & m) {
    for (int i = 0; i <= ones; i++) {
      if ((j == (1 << NN) - 1 && i == ones) || !solve(ss ^ m, i)) continue;
      int all = ((1 << NN) - 1) ^ j;
      for (int k = all;; k = k - 1 & all) {
        for (int a = i; a <= ones; a++) {
          if (DP[k][a - i]) {
            DP[k | j][a] = true;
          }
        }
        if (!k) break;
      }
    }
    if (!ss) break;
  }
  bool res = DP[(1 << NN) - 1][ones];
  memo[m | 1 << rt][ones] = res ? 1 : 0;
  return res;
}
int main() {
  cin >> N;
  int ones = 0;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    if (A[i] == 1) {
      ones++;
      N--;
      i--;
    }
  }
  memset(memo, -1, sizeof(memo));
  if (ones < N) {
    cout << "NO" << endl;
  } else if (solve((1 << N) - 1, ones)) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
int A[50];
int memo[1 << 12][30];
bool solve(int m, int ones) {
  if (m == 0) return ones == 1;
  if (memo[m][ones] != -1) return memo[m][ones] == 1;
  int rt = -1;
  int NN = __builtin_popcount(m) + ones;
  for (int i = 0; i < N; i++) {
    if (m & 1 << i) {
      if (A[i] == NN) {
        rt = i;
      }
    }
  }
  if (rt == -1) return false;
  m ^= 1 << rt;
  NN -= ones + 1;
  vector<vector<bool> > DP(1 << NN, vector<bool>(30, false));
  DP[0][0] = true;
  for (int j = 0, ss = m; j < 1 << NN; j++, ss = ss - 1 & m) {
    for (int i = 0; i <= ones; i++) {
      if ((j == (1 << NN) - 1 && i == ones) || !solve(ss ^ m, i)) continue;
      int all = ((1 << NN) - 1) ^ j;
      for (int k = all;; k = k - 1 & all) {
        for (int a = i; a <= ones; a++) {
          if (DP[k][a - i]) {
            DP[k | j][a] = true;
          }
        }
        if (!k) break;
      }
    }
    if (!ss) break;
  }
  bool res = DP[(1 << NN) - 1][ones];
  memo[m | 1 << rt][ones] = res ? 1 : 0;
  return res;
}
int main() {
  cin >> N;
  int ones = 0;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    if (A[i] == 1) {
      ones++;
      N--;
      i--;
    }
  }
  memset(memo, -1, sizeof(memo));
  if (ones < N) {
    cout << "NO" << endl;
  } else if (solve((1 << N) - 1, ones)) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}

```