#include <bits/stdc++.h>
using namespace std;
vector<int> nodes;
vector<int> toeat;
int used[(1 << 24)];
int n;
int r = 0;
void eat(int bm, int v, int tv, int x, int eated, int high, int ed,
         int skipped) {
  if (v > tv) {
    return;
  }
  if (tv == v && eated > 1) {
    toeat.push_back(bm);
    return;
  }
  for (int i = x + 1; i < n; i++) {
    if (v + nodes[i] > tv) return;
    if (bm & (1 << i)) {
      int nbm = bm - (1 << i);
      eat(nbm, v + nodes[i], tv, i, eated + 1, high, nodes[i], 0);
      eat(bm, v, tv, i, eated, high, nodes[i], 1);
      break;
    }
  }
}
void rec(int has, int x) {
  if (used[has]) return;
  if (r) return;
  if (x == n) {
    cout << "YES" << endl;
    r = 1;
    return;
  }
  toeat.clear();
  eat(has, 0, nodes[x] - 1, -1, 0, x, 0, 0);
  for (int l : toeat) {
    rec(l | (1 << x), x + 1);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    nodes.push_back(a);
    if (a == 2) {
      cout << "NO" << endl;
      return 0;
    }
  }
  sort(nodes.begin(), nodes.end());
  if (nodes[0] != 1) {
    cout << "NO" << endl;
    return 0;
  }
  if (nodes[n - 1] != n) {
    cout << "NO" << endl;
    return 0;
  }
  int init = 0;
  int sg = 0;
  for (int i = 0; i < n; i++) {
    if (nodes[i] == 1) {
      init |= (1 << i);
      sg = i;
    }
  }
  rec(init, sg + 1);
  if (!r) cout << "NO" << endl;
}

### 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> nodes;
vector<int> toeat;
int used[(1 << 24)];
int n;
int r = 0;
void eat(int bm, int v, int tv, int x, int eated, int high, int ed,
         int skipped) {
  if (v > tv) {
    return;
  }
  if (tv == v && eated > 1) {
    toeat.push_back(bm);
    return;
  }
  for (int i = x + 1; i < n; i++) {
    if (v + nodes[i] > tv) return;
    if (bm & (1 << i)) {
      int nbm = bm - (1 << i);
      eat(nbm, v + nodes[i], tv, i, eated + 1, high, nodes[i], 0);
      eat(bm, v, tv, i, eated, high, nodes[i], 1);
      break;
    }
  }
}
void rec(int has, int x) {
  if (used[has]) return;
  if (r) return;
  if (x == n) {
    cout << "YES" << endl;
    r = 1;
    return;
  }
  toeat.clear();
  eat(has, 0, nodes[x] - 1, -1, 0, x, 0, 0);
  for (int l : toeat) {
    rec(l | (1 << x), x + 1);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    nodes.push_back(a);
    if (a == 2) {
      cout << "NO" << endl;
      return 0;
    }
  }
  sort(nodes.begin(), nodes.end());
  if (nodes[0] != 1) {
    cout << "NO" << endl;
    return 0;
  }
  if (nodes[n - 1] != n) {
    cout << "NO" << endl;
    return 0;
  }
  int init = 0;
  int sg = 0;
  for (int i = 0; i < n; i++) {
    if (nodes[i] == 1) {
      init |= (1 << i);
      sg = i;
    }
  }
  rec(init, sg + 1);
  if (!r) cout << "NO" << endl;
}

```