#include <bits/stdc++.h>
using namespace std;
vector<int> a;
int vis[24][1 << 12], cnt[1 << 12], val[1 << 12];
int n;
bool rec(int node, int ones, int mask) {
  if (vis[ones][mask]) return false;
  vis[ones][mask] = true;
  if (node == n) return (mask == 1 << (n - 1)) && (ones == 0);
  for (int one = 0; one <= ones; one++) {
    for (int submask = mask;; submask = mask & (submask - 1)) {
      int no = val[submask];
      if (no + one + 1 == a[node] && one + cnt[submask] >= 2) {
        if (rec(node + 1, ones - one, (mask ^ submask) | (1 << node)))
          return true;
      }
      if (submask == 0) break;
    }
  }
  return false;
}
bool setup() {
  for (int mask = 0; mask < (1 << n); mask++) {
    int sum = 0, c = 0;
    for (int bit = 0; bit < n; bit++) {
      if (mask & (1 << bit)) c++, sum += a[bit];
    }
    val[mask] = sum, cnt[mask] = c;
  }
  return false;
}
int main() {
  int aa, ones = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> aa;
    if (aa == 1)
      ones++;
    else
      a.push_back(aa);
  }
  n -= ones;
  if (n == 0) {
    if (ones == 1)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
    return 0;
  }
  sort(a.begin(), a.end());
  bool answer = true;
  for (int i = 0; i < n; i++)
    if (a[i] == 2) answer = false;
  if ((ones < n) || n + ones != a[n - 1]) answer = false;
  if (!answer || setup() || !rec(0, ones, 0))
    cout << "NO" << endl;
  else
    cout << "YES" << 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;
vector<int> a;
int vis[24][1 << 12], cnt[1 << 12], val[1 << 12];
int n;
bool rec(int node, int ones, int mask) {
  if (vis[ones][mask]) return false;
  vis[ones][mask] = true;
  if (node == n) return (mask == 1 << (n - 1)) && (ones == 0);
  for (int one = 0; one <= ones; one++) {
    for (int submask = mask;; submask = mask & (submask - 1)) {
      int no = val[submask];
      if (no + one + 1 == a[node] && one + cnt[submask] >= 2) {
        if (rec(node + 1, ones - one, (mask ^ submask) | (1 << node)))
          return true;
      }
      if (submask == 0) break;
    }
  }
  return false;
}
bool setup() {
  for (int mask = 0; mask < (1 << n); mask++) {
    int sum = 0, c = 0;
    for (int bit = 0; bit < n; bit++) {
      if (mask & (1 << bit)) c++, sum += a[bit];
    }
    val[mask] = sum, cnt[mask] = c;
  }
  return false;
}
int main() {
  int aa, ones = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> aa;
    if (aa == 1)
      ones++;
    else
      a.push_back(aa);
  }
  n -= ones;
  if (n == 0) {
    if (ones == 1)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
    return 0;
  }
  sort(a.begin(), a.end());
  bool answer = true;
  for (int i = 0; i < n; i++)
    if (a[i] == 2) answer = false;
  if ((ones < n) || n + ones != a[n - 1]) answer = false;
  if (!answer || setup() || !rec(0, ones, 0))
    cout << "NO" << endl;
  else
    cout << "YES" << endl;
  return 0;
}

```