#include <bits/stdc++.h>
using namespace std;
vector<int> noleaves;
int sz;
int dp[15][1 << 15][15];
int dp2[1 << 15];
int solve(int node, int bitmask, int leaves) {
  if (leaves < 0) return 0;
  if (node == sz && !leaves) return 1;
  if (node == sz) return 0;
  if (!(bitmask & (1 << node))) return 0;
  if (dp[node][bitmask][leaves] != -1) return dp[node][bitmask][leaves];
  bool ans = false;
  int reversemask = (1 << sz) - 1;
  reversemask = bitmask ^ reversemask;
  for (int i = reversemask; i >= 0 && !ans; i--) {
    i &= reversemask;
    int cur = noleaves[node] - dp2[i] - 1;
    if (cur < 0) continue;
    int counter = cur;
    for (int j = 0; j < sz; j++) {
      if (i & (1 << j)) counter++;
    }
    if (counter < 2) continue;
    ans = ans || solve(node + 1, bitmask | i, leaves - cur);
  }
  return dp[node][bitmask][leaves] = ans;
}
int v[50];
int main() {
  int n;
  cin >> n;
  int hojas = 0;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    if (n == 1 && v[i] == 1) {
      cout << "YES" << endl;
      return 0;
    }
    if (v[i] == 1)
      hojas++;
    else
      noleaves.push_back(v[i]);
    if (v[i] == 2) {
      cout << "NO" << endl;
      return 0;
    }
  }
  sort(noleaves.begin(), noleaves.end());
  reverse(noleaves.begin(), noleaves.end());
  sz = noleaves.size();
  if (sz > 14) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < (1 << sz); i++) {
    int cur = 0;
    for (int j = 0; j < sz; j++)
      if (i & (1 << j)) cur += noleaves[j];
    dp2[i] = cur;
  }
  memset(dp, -1, sizeof dp);
  if (solve(0, 1, hojas))
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  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;
vector<int> noleaves;
int sz;
int dp[15][1 << 15][15];
int dp2[1 << 15];
int solve(int node, int bitmask, int leaves) {
  if (leaves < 0) return 0;
  if (node == sz && !leaves) return 1;
  if (node == sz) return 0;
  if (!(bitmask & (1 << node))) return 0;
  if (dp[node][bitmask][leaves] != -1) return dp[node][bitmask][leaves];
  bool ans = false;
  int reversemask = (1 << sz) - 1;
  reversemask = bitmask ^ reversemask;
  for (int i = reversemask; i >= 0 && !ans; i--) {
    i &= reversemask;
    int cur = noleaves[node] - dp2[i] - 1;
    if (cur < 0) continue;
    int counter = cur;
    for (int j = 0; j < sz; j++) {
      if (i & (1 << j)) counter++;
    }
    if (counter < 2) continue;
    ans = ans || solve(node + 1, bitmask | i, leaves - cur);
  }
  return dp[node][bitmask][leaves] = ans;
}
int v[50];
int main() {
  int n;
  cin >> n;
  int hojas = 0;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    if (n == 1 && v[i] == 1) {
      cout << "YES" << endl;
      return 0;
    }
    if (v[i] == 1)
      hojas++;
    else
      noleaves.push_back(v[i]);
    if (v[i] == 2) {
      cout << "NO" << endl;
      return 0;
    }
  }
  sort(noleaves.begin(), noleaves.end());
  reverse(noleaves.begin(), noleaves.end());
  sz = noleaves.size();
  if (sz > 14) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < (1 << sz); i++) {
    int cur = 0;
    for (int j = 0; j < sz; j++)
      if (i & (1 << j)) cur += noleaves[j];
    dp2[i] = cur;
  }
  memset(dp, -1, sizeof dp);
  if (solve(0, 1, hojas))
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

```