#include <bits/stdc++.h>
using namespace std;
int n, x, t, cnt, a[15], f[15][4200][15], size[4200], num[4200];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> x;
    if (x == 1)
      cnt++;
    else
      a[++t] = x;
  }
  for (int i = 1; i < t; ++i)
    for (int j = i + 1; j <= t; ++j)
      if (a[i] < a[j]) {
        int t = a[i];
        a[i] = a[j];
        a[j] = t;
      }
  if (n == 1 && cnt == 1)
    cout << "YES\n";
  else if (cnt <= n / 2 || cnt == n)
    cout << "NO\n";
  else {
    for (int i = 0; i < (1 << t); ++i)
      for (int j = 1; j <= t; ++j)
        if (i & (1 << (j - 1))) {
          size[i] += a[j];
          num[i]++;
        }
    f[0][1][cnt] = 1;
    for (int i = 0; i < t; ++i)
      for (int j = 0; j < (1 << t); ++j)
        for (int k = 0; k <= cnt; ++k)
          if (f[i][j][k]) {
            int p;
            for (int l = 0; l < (1 << t); ++l)
              if (!(j & l)) {
                p = a[i + 1] - size[l] - 1;
                if (p >= 0 && p <= k && p + num[l] > 1)
                  f[i + 1][j | l][k - p] = 1;
              }
          }
    if (f[t][(1 << t) - 1][0])
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  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, x, t, cnt, a[15], f[15][4200][15], size[4200], num[4200];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> x;
    if (x == 1)
      cnt++;
    else
      a[++t] = x;
  }
  for (int i = 1; i < t; ++i)
    for (int j = i + 1; j <= t; ++j)
      if (a[i] < a[j]) {
        int t = a[i];
        a[i] = a[j];
        a[j] = t;
      }
  if (n == 1 && cnt == 1)
    cout << "YES\n";
  else if (cnt <= n / 2 || cnt == n)
    cout << "NO\n";
  else {
    for (int i = 0; i < (1 << t); ++i)
      for (int j = 1; j <= t; ++j)
        if (i & (1 << (j - 1))) {
          size[i] += a[j];
          num[i]++;
        }
    f[0][1][cnt] = 1;
    for (int i = 0; i < t; ++i)
      for (int j = 0; j < (1 << t); ++j)
        for (int k = 0; k <= cnt; ++k)
          if (f[i][j][k]) {
            int p;
            for (int l = 0; l < (1 << t); ++l)
              if (!(j & l)) {
                p = a[i + 1] - size[l] - 1;
                if (p >= 0 && p <= k && p + num[l] > 1)
                  f[i + 1][j | l][k - p] = 1;
              }
          }
    if (f[t][(1 << t) - 1][0])
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}

```