#include <bits/stdc++.h>
using namespace std;
int n, k;
int a[100100];
int dp[13][(1 << 13) + 2][25];
bool f(int i1, int mask, int r) {
  if (i1 == k) {
    int b = 0;
    for (int j = 0; j < k; j++) {
      if (mask & (1 << j)) b++;
    }
    return (!r && b == 1) || (b == 0 && r == 1);
  }
  if (dp[i1][mask][r]) return 0;
  dp[i1][mask][r] = 1;
  int z[24];
  int k1 = 0;
  for (int i = 0; i < k; i++) {
    if (mask & (1 << i)) z[k1++] = i;
  }
  for (int i = 0; i < (1 << k1); i++) {
    int t = 0;
    int a1 = a[i1] - 1, r1 = 0;
    for (int j = 0; j < k1; j++) {
      if (i & (1 << j)) {
        a1 -= a[z[j]];
        r1 |= (1 << z[j]);
        t++;
      }
    }
    if (a1 < 0) continue;
    if (a1 > r) continue;
    int m1 = mask - r1;
    if (t + a1 == 1) continue;
    if (f(i1 + 1, m1 | (1 << i1), r - a1)) return 1;
  }
  return 0;
}
int main() {
  cin >> n;
  int b;
  for (int i = 0; i < n; i++) {
    cin >> b;
    if (b > 1) {
      a[k++] = b;
    }
    if (b == 2) {
      cout << "NO";
      return 0;
    }
  }
  if (k > 13) {
    cout << "NO";
    return 0;
  }
  sort(a, a + k);
  if (f(0, 0, n - k))
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

### Prompt

Please formulate 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, k;
int a[100100];
int dp[13][(1 << 13) + 2][25];
bool f(int i1, int mask, int r) {
  if (i1 == k) {
    int b = 0;
    for (int j = 0; j < k; j++) {
      if (mask & (1 << j)) b++;
    }
    return (!r && b == 1) || (b == 0 && r == 1);
  }
  if (dp[i1][mask][r]) return 0;
  dp[i1][mask][r] = 1;
  int z[24];
  int k1 = 0;
  for (int i = 0; i < k; i++) {
    if (mask & (1 << i)) z[k1++] = i;
  }
  for (int i = 0; i < (1 << k1); i++) {
    int t = 0;
    int a1 = a[i1] - 1, r1 = 0;
    for (int j = 0; j < k1; j++) {
      if (i & (1 << j)) {
        a1 -= a[z[j]];
        r1 |= (1 << z[j]);
        t++;
      }
    }
    if (a1 < 0) continue;
    if (a1 > r) continue;
    int m1 = mask - r1;
    if (t + a1 == 1) continue;
    if (f(i1 + 1, m1 | (1 << i1), r - a1)) return 1;
  }
  return 0;
}
int main() {
  cin >> n;
  int b;
  for (int i = 0; i < n; i++) {
    cin >> b;
    if (b > 1) {
      a[k++] = b;
    }
    if (b == 2) {
      cout << "NO";
      return 0;
    }
  }
  if (k > 13) {
    cout << "NO";
    return 0;
  }
  sort(a, a + k);
  if (f(0, 0, n - k))
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

```