#include <bits/stdc++.h>
using namespace std;
int mx[1 << 24];
int sum[1 << 24];
int a[25];
int lowbit(int i) { return i & (-i); }
bool flag[1 << 24];
vector<int> v[1005];
queue<int> q;
int main() {
  int i, j, k;
  int n;
  scanf("%d", &n);
  memset(sum, 0, sizeof(sum));
  memset(mx, 0, sizeof(mx));
  for (i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    mx[1 << i] = i + 1;
  }
  sort(a, a + n);
  for (i = 0; i < n; ++i) sum[1 << i] = a[i];
  if (n != a[n - 1]) {
    puts("NO");
    return 0;
  }
  v[0].push_back(0);
  for (i = 0; i < (1 << n); ++i) {
    if (i) mx[i] = max(mx[i - 1], mx[i]);
    if (i != 0 && sum[i] == 0) {
      sum[i] = sum[i ^ lowbit(i)] + sum[lowbit(i)];
      v[sum[i]].push_back(i);
    }
  }
  memset(flag, 0, sizeof(flag));
  q.push(1);
  flag[1] = true;
  int mm = (1 << (n - 1));
  while (!q.empty()) {
    i = q.front();
    q.pop();
    if (i & mm) continue;
    j = mx[i];
    for (k = 0; k < v[a[j] - 1].size(); ++k) {
      if ((i & (v[a[j] - 1][k])) != v[a[j] - 1][k]) continue;
      int ii = ((i ^ v[a[j] - 1][k]) | (1 << j));
      if (!flag[ii]) {
        flag[ii] = true;
        q.push(ii);
      }
    }
  }
  if (flag[mm])
    puts("YES");
  else
    puts("NO");
}

### 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 mx[1 << 24];
int sum[1 << 24];
int a[25];
int lowbit(int i) { return i & (-i); }
bool flag[1 << 24];
vector<int> v[1005];
queue<int> q;
int main() {
  int i, j, k;
  int n;
  scanf("%d", &n);
  memset(sum, 0, sizeof(sum));
  memset(mx, 0, sizeof(mx));
  for (i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    mx[1 << i] = i + 1;
  }
  sort(a, a + n);
  for (i = 0; i < n; ++i) sum[1 << i] = a[i];
  if (n != a[n - 1]) {
    puts("NO");
    return 0;
  }
  v[0].push_back(0);
  for (i = 0; i < (1 << n); ++i) {
    if (i) mx[i] = max(mx[i - 1], mx[i]);
    if (i != 0 && sum[i] == 0) {
      sum[i] = sum[i ^ lowbit(i)] + sum[lowbit(i)];
      v[sum[i]].push_back(i);
    }
  }
  memset(flag, 0, sizeof(flag));
  q.push(1);
  flag[1] = true;
  int mm = (1 << (n - 1));
  while (!q.empty()) {
    i = q.front();
    q.pop();
    if (i & mm) continue;
    j = mx[i];
    for (k = 0; k < v[a[j] - 1].size(); ++k) {
      if ((i & (v[a[j] - 1][k])) != v[a[j] - 1][k]) continue;
      int ii = ((i ^ v[a[j] - 1][k]) | (1 << j));
      if (!flag[ii]) {
        flag[ii] = true;
        q.push(ii);
      }
    }
  }
  if (flag[mm])
    puts("YES");
  else
    puts("NO");
}

```