#include <bits/stdc++.h>
using namespace std;
int c[24];
int n, leafcnt, intern;
bool cache[12][1 << 12][24];
void dp(int index, int visited, int leafused) {
  if (index == intern) {
    if (leafused == leafcnt) {
      printf("YES");
      exit(0);
    }
    return;
  }
  if (cache[index][visited][leafused]) return;
  cache[index][visited][leafused] = true;
  if (~visited & (1 << index)) {
    return;
  }
  int child, tmp, sum, i, cnt;
  for (child = 0; child < (1 << intern); child++) {
    if (visited & child) continue;
    for (i = 0, tmp = child, sum = 0, cnt = 0; tmp; i++, tmp >>= 1) {
      if (tmp & 1) {
        sum += c[i];
        cnt++;
      }
    }
    if (c[index] >= sum + 1 && cnt + c[index] - sum - 1 >= 2 &&
        leafused + c[index] - sum - 1 <= leafcnt) {
      dp(index + 1, visited | child, leafused + c[index] - sum - 1);
    }
  }
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", c + i);
    if (c[i] == 1) leafcnt++;
  }
  if (n + 1 > 2 * leafcnt) {
    printf("NO");
    return 0;
  }
  intern = n - leafcnt;
  sort(c, c + n, greater<int>());
  dp(0, 1, c[0] == 1);
  printf("NO");
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 c[24];
int n, leafcnt, intern;
bool cache[12][1 << 12][24];
void dp(int index, int visited, int leafused) {
  if (index == intern) {
    if (leafused == leafcnt) {
      printf("YES");
      exit(0);
    }
    return;
  }
  if (cache[index][visited][leafused]) return;
  cache[index][visited][leafused] = true;
  if (~visited & (1 << index)) {
    return;
  }
  int child, tmp, sum, i, cnt;
  for (child = 0; child < (1 << intern); child++) {
    if (visited & child) continue;
    for (i = 0, tmp = child, sum = 0, cnt = 0; tmp; i++, tmp >>= 1) {
      if (tmp & 1) {
        sum += c[i];
        cnt++;
      }
    }
    if (c[index] >= sum + 1 && cnt + c[index] - sum - 1 >= 2 &&
        leafused + c[index] - sum - 1 <= leafcnt) {
      dp(index + 1, visited | child, leafused + c[index] - sum - 1);
    }
  }
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", c + i);
    if (c[i] == 1) leafcnt++;
  }
  if (n + 1 > 2 * leafcnt) {
    printf("NO");
    return 0;
  }
  intern = n - leafcnt;
  sort(c, c + n, greater<int>());
  dp(0, 1, c[0] == 1);
  printf("NO");
  return 0;
}

```