#include <bits/stdc++.h>
using namespace std;
int n, m;
int a[24];
int child[24];
int lim[24];
void go(int i) {
  for (int j = i + 1; j < n; j++)
    if (!a[j] && child[j] < 2) return;
  if (a[i] <= 1) {
    if (a[i] == 1) {
      printf("NO");
      exit(0);
    } else {
      int need = 0;
      for (int j = i + 1; j < n; j++) need += a[j];
      if (need == m) {
        printf("YES");
        exit(0);
      } else {
        printf("NO");
        exit(0);
      }
    }
    return;
  }
  int _lim = lim[a[i]];
  for (int j = i + 1; j <= _lim; j++)
    if (a[j] > a[i]) {
      lim[a[i]] = j;
      a[j] -= a[i] + 1;
      child[j]++;
      if (a[j] || child[j] >= 2) go(i - 1);
      a[j] += a[i] + 1;
      child[j]--;
    }
  lim[a[i]] = _lim;
}
int main() {
  scanf("%d", &n);
  m = 0;
  for (int i = 0; i < n; i++) {
    lim[i] = n - 1;
    scanf("%d", a + i);
    a[i]--;
    m += a[i] == 0;
  }
  sort(a, a + n);
  if (n == 1) {
    printf("%s", a[0] == 0 ? "YES" : "NO");
    return 0;
  }
  if (n == 2) {
    printf("NO");
    return 0;
  }
  go(n - 2);
  printf("NO");
  return 0;
}

### Prompt

Your task is to create 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, m;
int a[24];
int child[24];
int lim[24];
void go(int i) {
  for (int j = i + 1; j < n; j++)
    if (!a[j] && child[j] < 2) return;
  if (a[i] <= 1) {
    if (a[i] == 1) {
      printf("NO");
      exit(0);
    } else {
      int need = 0;
      for (int j = i + 1; j < n; j++) need += a[j];
      if (need == m) {
        printf("YES");
        exit(0);
      } else {
        printf("NO");
        exit(0);
      }
    }
    return;
  }
  int _lim = lim[a[i]];
  for (int j = i + 1; j <= _lim; j++)
    if (a[j] > a[i]) {
      lim[a[i]] = j;
      a[j] -= a[i] + 1;
      child[j]++;
      if (a[j] || child[j] >= 2) go(i - 1);
      a[j] += a[i] + 1;
      child[j]--;
    }
  lim[a[i]] = _lim;
}
int main() {
  scanf("%d", &n);
  m = 0;
  for (int i = 0; i < n; i++) {
    lim[i] = n - 1;
    scanf("%d", a + i);
    a[i]--;
    m += a[i] == 0;
  }
  sort(a, a + n);
  if (n == 1) {
    printf("%s", a[0] == 0 ? "YES" : "NO");
    return 0;
  }
  if (n == 2) {
    printf("NO");
    return 0;
  }
  go(n - 2);
  printf("NO");
  return 0;
}

```