#include <bits/stdc++.h>
using namespace std;
long long n;
vector<long long> arr;
vector<long long> par;
vector<bool> hasParent;
long long memo[13][(1 << 14)][25] = {0};
long long solve(long long node, long long mask, long long leavesLeft) {
  if (arr[node] == 1) {
    if (leavesLeft == 0) {
      return 1;
    }
    return 2;
  }
  if (!hasParent[node]) return 2;
  if (memo[node][mask][leavesLeft]) {
    return memo[node][mask][leavesLeft];
  }
  for (long long sub = mask;; sub = (sub - 1) & mask) {
    long long bitCnt = 0;
    long long sumDegrees = 0;
    for (long long bit = 0; bit < n; ++bit) {
      if (sub & (1 << bit)) {
        bitCnt++;
        sumDegrees += arr[bit];
      }
    }
    if (sumDegrees < arr[node]) {
      for (long long bit = 0; bit < n; ++bit) {
        if (sub & (1 << bit)) {
          par[bit] = node;
          hasParent[bit] = true;
        }
      }
      long long remainingSpots = arr[node] - 1 - sumDegrees;
      long long children = bitCnt + remainingSpots;
      if (remainingSpots >= 0 && children >= 2 &&
          remainingSpots <= leavesLeft &&
          solve(node + 1, (mask ^ sub), leavesLeft - remainingSpots) == 1) {
        return memo[node][mask][leavesLeft] = 1;
      }
      for (long long bit = 0; bit < n; ++bit) {
        if (sub & (1 << bit)) {
          hasParent[bit] = false;
        }
      }
    }
    if (sub == 0) break;
  }
  return memo[node][mask][leavesLeft] = 2;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  if (n == 1) {
    cout << "YES\n";
    return 0;
  }
  arr.resize(n);
  hasParent.resize(n, false);
  par.resize(n);
  hasParent[0] = true;
  long long leafCnt = 0;
  for (long long i = 0; i < n; ++i) {
    cin >> arr[i];
    if (arr[i] == 2) {
      cout << "NO\n";
      return 0;
    }
    leafCnt += (arr[i] == 1);
  }
  if (leafCnt < (n + 1) / 2) {
    cout << "NO\n";
    return 0;
  }
  sort(arr.rbegin(), arr.rend());
  if (arr[0] != n) {
    cout << "NO\n";
    return 0;
  }
  n -= leafCnt;
  long long mask = (1 << n) - 2;
  if (solve(0, mask, leafCnt) == 1) {
    cout << "YES\n";
  } else {
    cout << "NO\n";
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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;
long long n;
vector<long long> arr;
vector<long long> par;
vector<bool> hasParent;
long long memo[13][(1 << 14)][25] = {0};
long long solve(long long node, long long mask, long long leavesLeft) {
  if (arr[node] == 1) {
    if (leavesLeft == 0) {
      return 1;
    }
    return 2;
  }
  if (!hasParent[node]) return 2;
  if (memo[node][mask][leavesLeft]) {
    return memo[node][mask][leavesLeft];
  }
  for (long long sub = mask;; sub = (sub - 1) & mask) {
    long long bitCnt = 0;
    long long sumDegrees = 0;
    for (long long bit = 0; bit < n; ++bit) {
      if (sub & (1 << bit)) {
        bitCnt++;
        sumDegrees += arr[bit];
      }
    }
    if (sumDegrees < arr[node]) {
      for (long long bit = 0; bit < n; ++bit) {
        if (sub & (1 << bit)) {
          par[bit] = node;
          hasParent[bit] = true;
        }
      }
      long long remainingSpots = arr[node] - 1 - sumDegrees;
      long long children = bitCnt + remainingSpots;
      if (remainingSpots >= 0 && children >= 2 &&
          remainingSpots <= leavesLeft &&
          solve(node + 1, (mask ^ sub), leavesLeft - remainingSpots) == 1) {
        return memo[node][mask][leavesLeft] = 1;
      }
      for (long long bit = 0; bit < n; ++bit) {
        if (sub & (1 << bit)) {
          hasParent[bit] = false;
        }
      }
    }
    if (sub == 0) break;
  }
  return memo[node][mask][leavesLeft] = 2;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  if (n == 1) {
    cout << "YES\n";
    return 0;
  }
  arr.resize(n);
  hasParent.resize(n, false);
  par.resize(n);
  hasParent[0] = true;
  long long leafCnt = 0;
  for (long long i = 0; i < n; ++i) {
    cin >> arr[i];
    if (arr[i] == 2) {
      cout << "NO\n";
      return 0;
    }
    leafCnt += (arr[i] == 1);
  }
  if (leafCnt < (n + 1) / 2) {
    cout << "NO\n";
    return 0;
  }
  sort(arr.rbegin(), arr.rend());
  if (arr[0] != n) {
    cout << "NO\n";
    return 0;
  }
  n -= leafCnt;
  long long mask = (1 << n) - 2;
  if (solve(0, mask, leafCnt) == 1) {
    cout << "YES\n";
  } else {
    cout << "NO\n";
  }
  return 0;
}

```