#include <bits/stdc++.h>
using namespace std;
int a[30];
int sum[10005];
int ones[10005], Log[10005];
bool dp[10005][30];
int main() {
  int n, m = 0, l = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int t;
    cin >> t;
    if (t == 1)
      l++;
    else
      a[m++] = t;
  }
  if (l < n / 2) {
    cout << "NO\n";
    return 0;
  }
  if (n == 1) {
    if (l == 1)
      cout << "YES\n";
    else
      cout << "NO\n";
    return 0;
  }
  sort(a, a + m);
  for (int i = 0; i < (1 << m); i++) {
    int s = 0;
    for (int j = 0; j < m; j++)
      if ((1 << j) & i) {
        s += a[j];
        ones[i]++;
        Log[i] = j;
      }
    sum[i] = s;
  }
  dp[0][l] = true;
  for (int i = 1; i < (1 << m); i++) {
    for (int j = 0; j <= l; j++) {
      int b = (1 << Log[i]);
      for (int k = 0; k < b; k++)
        if ((k & i) == 0 && sum[b] >= sum[k] + 1 &&
            ones[k] + (sum[b] - 1 - sum[k]) > 1)
          dp[i][j] |= dp[i + k - b][j + sum[b] - sum[k] - 1];
    }
    dp[0][l] = false;
  }
  if (dp[(1 << (m - 1))][0])
    cout << "YES\n";
  else
    cout << "NO\n";
  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 a[30];
int sum[10005];
int ones[10005], Log[10005];
bool dp[10005][30];
int main() {
  int n, m = 0, l = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int t;
    cin >> t;
    if (t == 1)
      l++;
    else
      a[m++] = t;
  }
  if (l < n / 2) {
    cout << "NO\n";
    return 0;
  }
  if (n == 1) {
    if (l == 1)
      cout << "YES\n";
    else
      cout << "NO\n";
    return 0;
  }
  sort(a, a + m);
  for (int i = 0; i < (1 << m); i++) {
    int s = 0;
    for (int j = 0; j < m; j++)
      if ((1 << j) & i) {
        s += a[j];
        ones[i]++;
        Log[i] = j;
      }
    sum[i] = s;
  }
  dp[0][l] = true;
  for (int i = 1; i < (1 << m); i++) {
    for (int j = 0; j <= l; j++) {
      int b = (1 << Log[i]);
      for (int k = 0; k < b; k++)
        if ((k & i) == 0 && sum[b] >= sum[k] + 1 &&
            ones[k] + (sum[b] - 1 - sum[k]) > 1)
          dp[i][j] |= dp[i + k - b][j + sum[b] - sum[k] - 1];
    }
    dp[0][l] = false;
  }
  if (dp[(1 << (m - 1))][0])
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}

```