#include <bits/stdc++.h>
using namespace std;
int N, a[24];
vector<int> dp(1 << 24, 0), tp(1 << 24, 0);
int co[1 << 24];
int n;
int main() {
  for (int i = 0; i < 24; i++) co[1 << i] = i;
  ios_base::sync_with_stdio(false);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> a[i];
  sort(a, a + N);
  n = (1 << N) - 1;
  bool tet = false;
  for (int i = 0; i < N; i++)
    if (a[i] == 2) tet = true;
  if (N == 1 && a[0] == 1) {
    cout << "YES" << endl;
    return 0;
  }
  if (a[0] != 1 || a[N - 1] != N || a[N - 2] == N - 1 || tet) {
    cout << "NO" << endl;
    return 0;
  }
  int p;
  for (int i = 0; i < N; i++)
    if (a[i] != 1) {
      p = i;
      tp[n] = 1;
      break;
    }
  for (int i = p; i < N; i++) {
    dp = tp;
    tp.assign(1 << 24, 0);
    for (int j = 0; j < (1 << N); j++) {
      if (!dp[j]) continue;
      int t = a[i] - 1;
      vector<int> ll;
      for (int k = 0; k < i; k++)
        if ((1 << k) & j) ll.push_back(k);
      int sz = ll.size();
      for (int k = 0; k < (1 << sz); k++) {
        int tot = 0, mask = 0;
        for (int l = k; l; l &= l - 1) {
          tot += a[ll[co[l & -l]]];
          mask |= 1 << ll[co[l & -l]];
        }
        if (tot == t && __builtin_popcount(mask) > 1) tp[j & ~mask] = 1;
      }
    }
  }
  if (tp[1 << (N - 1)])
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

### Prompt

Generate 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;
int N, a[24];
vector<int> dp(1 << 24, 0), tp(1 << 24, 0);
int co[1 << 24];
int n;
int main() {
  for (int i = 0; i < 24; i++) co[1 << i] = i;
  ios_base::sync_with_stdio(false);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> a[i];
  sort(a, a + N);
  n = (1 << N) - 1;
  bool tet = false;
  for (int i = 0; i < N; i++)
    if (a[i] == 2) tet = true;
  if (N == 1 && a[0] == 1) {
    cout << "YES" << endl;
    return 0;
  }
  if (a[0] != 1 || a[N - 1] != N || a[N - 2] == N - 1 || tet) {
    cout << "NO" << endl;
    return 0;
  }
  int p;
  for (int i = 0; i < N; i++)
    if (a[i] != 1) {
      p = i;
      tp[n] = 1;
      break;
    }
  for (int i = p; i < N; i++) {
    dp = tp;
    tp.assign(1 << 24, 0);
    for (int j = 0; j < (1 << N); j++) {
      if (!dp[j]) continue;
      int t = a[i] - 1;
      vector<int> ll;
      for (int k = 0; k < i; k++)
        if ((1 << k) & j) ll.push_back(k);
      int sz = ll.size();
      for (int k = 0; k < (1 << sz); k++) {
        int tot = 0, mask = 0;
        for (int l = k; l; l &= l - 1) {
          tot += a[ll[co[l & -l]]];
          mask |= 1 << ll[co[l & -l]];
        }
        if (tot == t && __builtin_popcount(mask) > 1) tp[j & ~mask] = 1;
      }
    }
  }
  if (tp[1 << (N - 1)])
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

```