#include <bits/stdc++.h>
using namespace std;
int n, m, ONE, a[20];
int f[1 << 11][25];
vector<pair<int, int> > s[1 << 11];
bool dp(int state, int nOne) {
  if (f[state][nOne] != -1) return f[state][nOne];
  f[state][nOne] = false;
  int id = -1;
  for (int i = 0; i < (n); i++)
    if (state & (1 << i)) {
      id = i;
      break;
    }
  if (id == -1) {
    if (nOne) return false;
    f[state][nOne] = true;
    return true;
  }
  int cur = a[id] - 1;
  for (int one = (0); one <= (min(cur, nOne)); one++)
    for (int i = 0; i < (s[cur - one].size()); i++) {
      int sub = s[cur - one][i].first;
      if (((sub & state) == 0) && (s[cur - one][i].second + one >= 2)) {
        int new_state = state ^ (1 << id) ^ sub;
        if (dp(new_state, nOne - one)) {
          f[state][nOne] = true;
          return true;
        }
      }
    }
  return false;
}
void solve() {
  if (n >= 12) {
    cout << "NO" << endl;
    return;
  }
  if (n == 0) {
    if (ONE == 1)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
    return;
  }
  for (int state = 0; state < ((1 << n)); state++) {
    int cnt = 0, total = 0;
    for (int i = 0; i < (n); i++)
      if (state & (1 << i)) cnt++, total += a[i];
    s[total].push_back(make_pair(state, cnt));
  }
  memset(f, -1, sizeof(f));
  if (dp(1, ONE) == 1)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
int main() {
  cin >> m;
  int u;
  for (int i = (1); i <= (m); i++) {
    cin >> u;
    if (u == 1)
      ONE++;
    else
      a[n++] = u;
  }
  for (int i = 0; i < (n); i++)
    for (int j = (i + 1); j <= (n - 1); j++)
      if (a[i] < a[j]) swap(a[i], a[j]);
  solve();
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int n, m, ONE, a[20];
int f[1 << 11][25];
vector<pair<int, int> > s[1 << 11];
bool dp(int state, int nOne) {
  if (f[state][nOne] != -1) return f[state][nOne];
  f[state][nOne] = false;
  int id = -1;
  for (int i = 0; i < (n); i++)
    if (state & (1 << i)) {
      id = i;
      break;
    }
  if (id == -1) {
    if (nOne) return false;
    f[state][nOne] = true;
    return true;
  }
  int cur = a[id] - 1;
  for (int one = (0); one <= (min(cur, nOne)); one++)
    for (int i = 0; i < (s[cur - one].size()); i++) {
      int sub = s[cur - one][i].first;
      if (((sub & state) == 0) && (s[cur - one][i].second + one >= 2)) {
        int new_state = state ^ (1 << id) ^ sub;
        if (dp(new_state, nOne - one)) {
          f[state][nOne] = true;
          return true;
        }
      }
    }
  return false;
}
void solve() {
  if (n >= 12) {
    cout << "NO" << endl;
    return;
  }
  if (n == 0) {
    if (ONE == 1)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
    return;
  }
  for (int state = 0; state < ((1 << n)); state++) {
    int cnt = 0, total = 0;
    for (int i = 0; i < (n); i++)
      if (state & (1 << i)) cnt++, total += a[i];
    s[total].push_back(make_pair(state, cnt));
  }
  memset(f, -1, sizeof(f));
  if (dp(1, ONE) == 1)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
int main() {
  cin >> m;
  int u;
  for (int i = (1); i <= (m); i++) {
    cin >> u;
    if (u == 1)
      ONE++;
    else
      a[n++] = u;
  }
  for (int i = 0; i < (n); i++)
    for (int j = (i + 1); j <= (n - 1); j++)
      if (a[i] < a[j]) swap(a[i], a[j]);
  solve();
  return 0;
}

```