#include <bits/stdc++.h>
using namespace std;
set<pair<int, int> > ve;
int num[10000], siz[10000], res, n, vis[10000], tot = 0, big = 0, xx = 0,
                                                l[10000];
int cur[10000];
set<pair<int, int> >::iterator it;
set<int> Set;
int dfs(int root, int cnt) {
  int i;
  if (root >= n - 1) {
    int sum = cnt;
    int go = 0;
    for (i = siz[1]; i < n - 1; i++) {
      go += cur[i] - 1;
      if (cur[i] == 1 && l[i] == 1) return 0;
    }
    if (xx == 1 && big - 1 - sum == 0) return 0;
    go += big - 1 - sum;
    if (big - 1 >= sum && go == siz[1]) return 1;
    return 0;
  }
  for (i = root + 1; i < n - 1; i++) {
    if (cur[i] > num[root]) {
      l[i]++;
      cur[i] -= num[root];
      if (dfs(root + 1, cnt)) return 1;
      cur[i] += num[root];
      l[i]--;
    }
  }
  xx++;
  cnt += num[root];
  if (dfs(root + 1, cnt)) return 1;
  cnt -= num[root];
  xx--;
  return 0;
}
int main() {
  memset(siz, 0, sizeof(siz));
  memset(vis, 0, sizeof(vis));
  memset(l, 0, sizeof(l));
  int i, j;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &num[i]), tot += num[i];
  for (i = 0; i < n; i++) cur[i] = num[i];
  for (i = 0; i < n; i++) siz[num[i]]++;
  if (siz[1] < (n + 1) / 2 || siz[2]) {
    printf("NO\n");
    return 0;
  }
  sort(num, num + n);
  sort(cur, cur + n);
  res = n - siz[1];
  big = num[n - 1];
  if (n == 1) {
    printf("YES\n");
    return 0;
  }
  if (dfs(siz[1], 0))
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
set<pair<int, int> > ve;
int num[10000], siz[10000], res, n, vis[10000], tot = 0, big = 0, xx = 0,
                                                l[10000];
int cur[10000];
set<pair<int, int> >::iterator it;
set<int> Set;
int dfs(int root, int cnt) {
  int i;
  if (root >= n - 1) {
    int sum = cnt;
    int go = 0;
    for (i = siz[1]; i < n - 1; i++) {
      go += cur[i] - 1;
      if (cur[i] == 1 && l[i] == 1) return 0;
    }
    if (xx == 1 && big - 1 - sum == 0) return 0;
    go += big - 1 - sum;
    if (big - 1 >= sum && go == siz[1]) return 1;
    return 0;
  }
  for (i = root + 1; i < n - 1; i++) {
    if (cur[i] > num[root]) {
      l[i]++;
      cur[i] -= num[root];
      if (dfs(root + 1, cnt)) return 1;
      cur[i] += num[root];
      l[i]--;
    }
  }
  xx++;
  cnt += num[root];
  if (dfs(root + 1, cnt)) return 1;
  cnt -= num[root];
  xx--;
  return 0;
}
int main() {
  memset(siz, 0, sizeof(siz));
  memset(vis, 0, sizeof(vis));
  memset(l, 0, sizeof(l));
  int i, j;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &num[i]), tot += num[i];
  for (i = 0; i < n; i++) cur[i] = num[i];
  for (i = 0; i < n; i++) siz[num[i]]++;
  if (siz[1] < (n + 1) / 2 || siz[2]) {
    printf("NO\n");
    return 0;
  }
  sort(num, num + n);
  sort(cur, cur + n);
  res = n - siz[1];
  big = num[n - 1];
  if (n == 1) {
    printf("YES\n");
    return 0;
  }
  if (dfs(siz[1], 0))
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}

```