#include <bits/stdc++.h>
using namespace std;
vector<int> q;
int stk[20];
int top;
int cnt;
int vis[20];
int dfs(int cent) {
  if (cent < 0) {
    if (cnt == 0)
      return 1;
    else
      return 0;
  }
  if (vis[cent] == 0) return 0;
  int mx = 1 << cent;
  for (int s = 0; s < mx; s++) {
    int an = 0;
    int res = 0;
    for (int i = 0; i < cent; i++) {
      if (vis[i]) continue;
      if ((1 << i) & s) {
        an += stk[i];
        res++;
      }
    }
    if (cnt + an >= stk[cent] - 1 && an <= stk[cent] - 1) {
      if (an == stk[cent] - 1 && res == 1 || res + cnt < 2) continue;
      int vis2[20];
      memset(vis2, 0, sizeof(vis2));
      cnt -= stk[cent] - 1 - an;
      for (int i = 0; i < cent; i++) {
        if (vis[i]) continue;
        if ((1 << i) & s) {
          vis2[i] = vis[i] = 1;
        }
      }
      int ok = dfs(cent - 1);
      cnt += stk[cent] - 1 - an;
      for (int i = 0; i < cent; i++) {
        if (vis2[i]) {
          vis2[i] = vis[i] = 0;
        }
      }
      if (ok == 1) return 1;
    }
  }
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int v;
    scanf("%d", &v);
    if (v == 2) {
      printf("NO\n");
      return 0;
    }
    if (v == 1)
      cnt++;
    else
      stk[top++] = v;
  }
  if (cnt < n / 2) {
    printf("NO\n");
    return 0;
  }
  sort(stk, stk + top);
  if (top == 0 && cnt == 1)
    printf("YES\n");
  else if (top == 0)
    printf("NO\n");
  else {
    vis[top - 1] = 1;
    int ok = dfs(top - 1);
    if (ok)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
vector<int> q;
int stk[20];
int top;
int cnt;
int vis[20];
int dfs(int cent) {
  if (cent < 0) {
    if (cnt == 0)
      return 1;
    else
      return 0;
  }
  if (vis[cent] == 0) return 0;
  int mx = 1 << cent;
  for (int s = 0; s < mx; s++) {
    int an = 0;
    int res = 0;
    for (int i = 0; i < cent; i++) {
      if (vis[i]) continue;
      if ((1 << i) & s) {
        an += stk[i];
        res++;
      }
    }
    if (cnt + an >= stk[cent] - 1 && an <= stk[cent] - 1) {
      if (an == stk[cent] - 1 && res == 1 || res + cnt < 2) continue;
      int vis2[20];
      memset(vis2, 0, sizeof(vis2));
      cnt -= stk[cent] - 1 - an;
      for (int i = 0; i < cent; i++) {
        if (vis[i]) continue;
        if ((1 << i) & s) {
          vis2[i] = vis[i] = 1;
        }
      }
      int ok = dfs(cent - 1);
      cnt += stk[cent] - 1 - an;
      for (int i = 0; i < cent; i++) {
        if (vis2[i]) {
          vis2[i] = vis[i] = 0;
        }
      }
      if (ok == 1) return 1;
    }
  }
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int v;
    scanf("%d", &v);
    if (v == 2) {
      printf("NO\n");
      return 0;
    }
    if (v == 1)
      cnt++;
    else
      stk[top++] = v;
  }
  if (cnt < n / 2) {
    printf("NO\n");
    return 0;
  }
  sort(stk, stk + top);
  if (top == 0 && cnt == 1)
    printf("YES\n");
  else if (top == 0)
    printf("NO\n");
  else {
    vis[top - 1] = 1;
    int ok = dfs(top - 1);
    if (ok)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

```