#include <bits/stdc++.h>
using namespace std;
int f1[10001], f2[10001], pre[10001], num[10001], n, sum;
int a[10001], use[10001];
int main() {
  scanf("%d", &n);
  if (n == 1) {
    puts("YES");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] > sum) sum = a[i];
    if (a[i] == 2) {
      puts("NO");
      return 0;
    }
  }
  if (sum != n) {
    puts("NO");
    return 0;
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) continue;
    for (int j = 0; j <= a[i]; j++) f1[j] = f2[j] = pre[j] = 0;
    for (int j = 1; j < i; j++) {
      if (use[j]) continue;
      for (int k = a[i] - a[j]; k >= 0; k--) {
        if (f1[k]) f2[k + a[j]] = 1, pre[k + a[j]] = j;
        if (f2[k])
          if (!f2[k + a[j]]) f2[k + a[j]] = 1, pre[k + a[j]] = j;
      }
      f1[a[j]] = 1;
      num[a[j]] = j;
    }
    if (!f2[a[i] - 1]) {
      puts("NO");
      return 0;
    }
    int tmp;
    for (tmp = a[i] - 1; pre[tmp]; tmp = tmp - a[pre[tmp]]) use[pre[tmp]] = 1;
    use[num[tmp]] = 1;
  }
  puts("YES");
}

### 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;
int f1[10001], f2[10001], pre[10001], num[10001], n, sum;
int a[10001], use[10001];
int main() {
  scanf("%d", &n);
  if (n == 1) {
    puts("YES");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] > sum) sum = a[i];
    if (a[i] == 2) {
      puts("NO");
      return 0;
    }
  }
  if (sum != n) {
    puts("NO");
    return 0;
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) continue;
    for (int j = 0; j <= a[i]; j++) f1[j] = f2[j] = pre[j] = 0;
    for (int j = 1; j < i; j++) {
      if (use[j]) continue;
      for (int k = a[i] - a[j]; k >= 0; k--) {
        if (f1[k]) f2[k + a[j]] = 1, pre[k + a[j]] = j;
        if (f2[k])
          if (!f2[k + a[j]]) f2[k + a[j]] = 1, pre[k + a[j]] = j;
      }
      f1[a[j]] = 1;
      num[a[j]] = j;
    }
    if (!f2[a[i] - 1]) {
      puts("NO");
      return 0;
    }
    int tmp;
    for (tmp = a[i] - 1; pre[tmp]; tmp = tmp - a[pre[tmp]]) use[pre[tmp]] = 1;
    use[num[tmp]] = 1;
  }
  puts("YES");
}

```