#include <bits/stdc++.h>
using namespace std;
const int N = 25;
vector<pair<int, int> > vc[1 << 13];
int M[1 << 13][25][25], A[25], H[1 << 13], m, n, bir, i, x, asd;
void g(int m, int i, int m2, int t) {
  if (i == n) {
    vc[m].push_back(make_pair(t, m2));
    return;
  }
  if (!(m & (1 << i))) g(m, i + 1, m2 | (1 << i), t + A[i]);
  g(m, i + 1, m2, t);
}
void f(int m, int i, int bir) {
  if (bir < 0 || asd) return;
  if (i == n) {
    if ((m == (1 << n - 1) - 1) && !bir) asd = 1;
    return;
  }
  if (M[m][i][bir]) return;
  M[m][i][bir] = 1;
  if (!H[m]) {
    H[m] = 1;
    vc[m].clear();
    g(m, 0, 0, 0);
  }
  for (int j = 0; j < vc[m].size(); j++)
    if (vc[m][j].first <= A[i] - 1) {
      if (__builtin_popcount(vc[m][j].second) + A[i] - 1 - vc[m][j].first == 1)
        continue;
      f(m | vc[m][j].second, i + 1, bir - (A[i] - 1 - vc[m][j].first));
    }
}
int main() {
  int w;
  cin >> n;
  bir = m = 0;
  for (i = 1; i <= n; i++) {
    cin >> x;
    if (x > 1)
      A[m++] = x;
    else
      bir++;
  }
  if (n == 1 && bir == 1) {
    cout << "YES\n" << endl;
    return 0;
  }
  n = m;
  if (n > 12) {
    cout << "NO\n" << endl;
    return 0;
  }
  sort(A, A + n);
  memset(H, asd = 0, sizeof H);
  f(0, 0, bir);
  cout << (asd ? "YES\n" : "NO\n");
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int N = 25;
vector<pair<int, int> > vc[1 << 13];
int M[1 << 13][25][25], A[25], H[1 << 13], m, n, bir, i, x, asd;
void g(int m, int i, int m2, int t) {
  if (i == n) {
    vc[m].push_back(make_pair(t, m2));
    return;
  }
  if (!(m & (1 << i))) g(m, i + 1, m2 | (1 << i), t + A[i]);
  g(m, i + 1, m2, t);
}
void f(int m, int i, int bir) {
  if (bir < 0 || asd) return;
  if (i == n) {
    if ((m == (1 << n - 1) - 1) && !bir) asd = 1;
    return;
  }
  if (M[m][i][bir]) return;
  M[m][i][bir] = 1;
  if (!H[m]) {
    H[m] = 1;
    vc[m].clear();
    g(m, 0, 0, 0);
  }
  for (int j = 0; j < vc[m].size(); j++)
    if (vc[m][j].first <= A[i] - 1) {
      if (__builtin_popcount(vc[m][j].second) + A[i] - 1 - vc[m][j].first == 1)
        continue;
      f(m | vc[m][j].second, i + 1, bir - (A[i] - 1 - vc[m][j].first));
    }
}
int main() {
  int w;
  cin >> n;
  bir = m = 0;
  for (i = 1; i <= n; i++) {
    cin >> x;
    if (x > 1)
      A[m++] = x;
    else
      bir++;
  }
  if (n == 1 && bir == 1) {
    cout << "YES\n" << endl;
    return 0;
  }
  n = m;
  if (n > 12) {
    cout << "NO\n" << endl;
    return 0;
  }
  sort(A, A + n);
  memset(H, asd = 0, sizeof H);
  f(0, 0, bir);
  cout << (asd ? "YES\n" : "NO\n");
  return 0;
}

```