#include <bits/stdc++.h>
using namespace std;
int N, SZ;
int arr[30];
short LOG[1 << 15];
int mi = 0;
vector<int> ADJ[25];
int SUM[1 << 14];
int mem[1 << 14][24];
int dp(int mask, int ones) {
  int b = mask & -mask;
  int id = LOG[b];
  if (mask == 0) return ones == 0;
  if (mem[mask][ones] != -1) return mem[mask][ones];
  int w = 0;
  for (int j = 0; j <= ones && j <= arr[id] - 1 && !w; ++j) {
    for (int i = 0; i < ADJ[arr[id] - j - 1].size() && !w; ++i) {
      int nm = ADJ[arr[id] - j - 1][i];
      if (nm & mask) continue;
      if (id + 1 != SZ && ((mask ^ nm) & (b << 1)) == 0) continue;
      if (j <= 1 && nm == 0) continue;
      if (j == 0 && ((nm & -nm) == nm)) continue;
      if (dp(mask ^ nm ^ b, ones - j)) w = 1;
    }
  }
  return mem[mask][ones] = w;
}
int main() {
  memset(mem, -1, sizeof(mem));
  cin >> N;
  for (int i = 0; i < 15; ++i) LOG[1 << i] = i;
  for (int i = 0; i < N; ++i) {
    cin >> arr[i];
  }
  if (N == 1 && arr[0] == 1) {
    cout << "YES" << endl;
    return 0;
  }
  sort(arr, arr + N);
  reverse(arr, arr + N);
  if (N != arr[0]) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < N; ++i) {
    SZ = i;
    if (arr[i] == 1) break;
  }
  if (arr[N - 1] != 1) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < 25; ++i) ADJ[i].reserve(500000);
  if (SZ >= 14) {
    cout << "NO" << endl;
    return 0;
  }
  ADJ[0].push_back(0);
  for (int i = 1; i < (1 << SZ); ++i) {
    int b = i & -i;
    SUM[i] = SUM[i ^ b] + arr[LOG[b]];
    if (SUM[i] < 25) ADJ[SUM[i]].push_back(i);
  }
  cout << (dp(1, N - SZ) ? "YES" : "NO") << endl;
  return 0;
}

### 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, SZ;
int arr[30];
short LOG[1 << 15];
int mi = 0;
vector<int> ADJ[25];
int SUM[1 << 14];
int mem[1 << 14][24];
int dp(int mask, int ones) {
  int b = mask & -mask;
  int id = LOG[b];
  if (mask == 0) return ones == 0;
  if (mem[mask][ones] != -1) return mem[mask][ones];
  int w = 0;
  for (int j = 0; j <= ones && j <= arr[id] - 1 && !w; ++j) {
    for (int i = 0; i < ADJ[arr[id] - j - 1].size() && !w; ++i) {
      int nm = ADJ[arr[id] - j - 1][i];
      if (nm & mask) continue;
      if (id + 1 != SZ && ((mask ^ nm) & (b << 1)) == 0) continue;
      if (j <= 1 && nm == 0) continue;
      if (j == 0 && ((nm & -nm) == nm)) continue;
      if (dp(mask ^ nm ^ b, ones - j)) w = 1;
    }
  }
  return mem[mask][ones] = w;
}
int main() {
  memset(mem, -1, sizeof(mem));
  cin >> N;
  for (int i = 0; i < 15; ++i) LOG[1 << i] = i;
  for (int i = 0; i < N; ++i) {
    cin >> arr[i];
  }
  if (N == 1 && arr[0] == 1) {
    cout << "YES" << endl;
    return 0;
  }
  sort(arr, arr + N);
  reverse(arr, arr + N);
  if (N != arr[0]) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < N; ++i) {
    SZ = i;
    if (arr[i] == 1) break;
  }
  if (arr[N - 1] != 1) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < 25; ++i) ADJ[i].reserve(500000);
  if (SZ >= 14) {
    cout << "NO" << endl;
    return 0;
  }
  ADJ[0].push_back(0);
  for (int i = 1; i < (1 << SZ); ++i) {
    int b = i & -i;
    SUM[i] = SUM[i ^ b] + arr[LOG[b]];
    if (SUM[i] < 25) ADJ[SUM[i]].push_back(i);
  }
  cout << (dp(1, N - SZ) ? "YES" : "NO") << endl;
  return 0;
}

```