#include <bits/stdc++.h>
using namespace std;
const int N = 30;
int n;
int cnt[N];
int used[N];
vector<int> items;
vector<int> remain;
vector<int> reqs;
void dfs2(int, int, int);
void dfs1(int i) {
  if (i == reqs.size()) {
    printf("YES\n");
    exit(0);
  }
  int mx = 0;
  for (int t = 0; t < items.size(); t++)
    if (remain[t]) mx = max(mx, items[t]);
  if (reqs[i] <= mx) return;
  dfs2(i, 0, reqs[i]);
}
void dfs2(int i, int j, int rem) {
  if (rem < 0) return;
  if (rem == 0) {
    dfs1(i + 1);
    return;
  }
  if (j == items.size()) return;
  int tot = remain[j];
  for (int c = tot; c >= 0; c--) {
    remain[j] -= c;
    dfs2(i, j + 1, rem - c * items[j]);
    remain[j] += c;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", cnt + i);
  int total = 0, sum = 0, c = 0;
  bool exist = false;
  for (int i = 1; i <= n; i++) total += cnt[i] - 1;
  for (int i = 1; i <= n; i++) {
    if (cnt[i] != n) {
      sum += cnt[i];
    } else
      c++;
    if (cnt[i] == 2) exist = true;
  }
  if (c == 1 && sum == total && !exist) {
    for (int i = 1; i <= n; i++) {
      if (cnt[i] != n) items.push_back(cnt[i]);
      if (cnt[i] != 1) reqs.push_back(cnt[i] - 1);
    }
    sort(items.begin(), items.end(), greater<int>());
    sort(reqs.begin(), reqs.end(), greater<int>());
    for (int p = 0, q; p < items.size(); p = q + 1) {
      for (q = p; q + 1 < items.size() && items[q + 1] == items[p]; q++)
        ;
      remain.push_back(q - p + 1);
    }
    items.erase(unique(items.begin(), items.end()), items.end());
    dfs1(0);
    printf("NO\n");
  } else {
    printf("NO\n");
  }
}

### Prompt

Your task is to create 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;
const int N = 30;
int n;
int cnt[N];
int used[N];
vector<int> items;
vector<int> remain;
vector<int> reqs;
void dfs2(int, int, int);
void dfs1(int i) {
  if (i == reqs.size()) {
    printf("YES\n");
    exit(0);
  }
  int mx = 0;
  for (int t = 0; t < items.size(); t++)
    if (remain[t]) mx = max(mx, items[t]);
  if (reqs[i] <= mx) return;
  dfs2(i, 0, reqs[i]);
}
void dfs2(int i, int j, int rem) {
  if (rem < 0) return;
  if (rem == 0) {
    dfs1(i + 1);
    return;
  }
  if (j == items.size()) return;
  int tot = remain[j];
  for (int c = tot; c >= 0; c--) {
    remain[j] -= c;
    dfs2(i, j + 1, rem - c * items[j]);
    remain[j] += c;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", cnt + i);
  int total = 0, sum = 0, c = 0;
  bool exist = false;
  for (int i = 1; i <= n; i++) total += cnt[i] - 1;
  for (int i = 1; i <= n; i++) {
    if (cnt[i] != n) {
      sum += cnt[i];
    } else
      c++;
    if (cnt[i] == 2) exist = true;
  }
  if (c == 1 && sum == total && !exist) {
    for (int i = 1; i <= n; i++) {
      if (cnt[i] != n) items.push_back(cnt[i]);
      if (cnt[i] != 1) reqs.push_back(cnt[i] - 1);
    }
    sort(items.begin(), items.end(), greater<int>());
    sort(reqs.begin(), reqs.end(), greater<int>());
    for (int p = 0, q; p < items.size(); p = q + 1) {
      for (q = p; q + 1 < items.size() && items[q + 1] == items[p]; q++)
        ;
      remain.push_back(q - p + 1);
    }
    items.erase(unique(items.begin(), items.end()), items.end());
    dfs1(0);
    printf("NO\n");
  } else {
    printf("NO\n");
  }
}

```