#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> t;
int a[100];
int h[100];
bool ok;
int y;
void haku(int k) {
  if (ok) return;
  if (t[k] == 1) {
    int s = 0;
    for (int i = 0; i < k; i++) s += h[i];
    if (s == y) ok = true;
    return;
  }
  for (int i = 0; i < k; i++) {
    if (a[i] == t[k]) continue;
    if (h[i] - t[k] < 0) continue;
    h[i] -= t[k];
    a[k] = t[k] - 1;
    h[k] = t[k] - 1;
    haku(k + 1);
    h[i] += t[k];
  }
}
int main() {
  cin >> n;
  t.resize(n);
  for (int i = 0; i < n; i++) cin >> t[i];
  for (int i = 0; i < n; i++) {
    if (t[i] == 1) y++;
    if (t[i] == 2) {
      cout << "NO\n";
      return 0;
    }
  }
  if (y * 2 < n) {
    cout << "NO\n";
    return 0;
  }
  sort(t.rbegin(), t.rend());
  a[0] = t[0] - 1;
  h[0] = t[0] - 1;
  haku(1);
  if (t[0] == 1 && n == 1) ok = true;
  if (ok)
    cout << "YES\n";
  else
    cout << "NO\n";
}

### Prompt

Develop a solution in Cpp 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;
vector<int> t;
int a[100];
int h[100];
bool ok;
int y;
void haku(int k) {
  if (ok) return;
  if (t[k] == 1) {
    int s = 0;
    for (int i = 0; i < k; i++) s += h[i];
    if (s == y) ok = true;
    return;
  }
  for (int i = 0; i < k; i++) {
    if (a[i] == t[k]) continue;
    if (h[i] - t[k] < 0) continue;
    h[i] -= t[k];
    a[k] = t[k] - 1;
    h[k] = t[k] - 1;
    haku(k + 1);
    h[i] += t[k];
  }
}
int main() {
  cin >> n;
  t.resize(n);
  for (int i = 0; i < n; i++) cin >> t[i];
  for (int i = 0; i < n; i++) {
    if (t[i] == 1) y++;
    if (t[i] == 2) {
      cout << "NO\n";
      return 0;
    }
  }
  if (y * 2 < n) {
    cout << "NO\n";
    return 0;
  }
  sort(t.rbegin(), t.rend());
  a[0] = t[0] - 1;
  h[0] = t[0] - 1;
  haku(1);
  if (t[0] == 1 && n == 1) ok = true;
  if (ok)
    cout << "YES\n";
  else
    cout << "NO\n";
}

```