#include <bits/stdc++.h>
using namespace std;
const int N = 24;
int n, a[N];
int dp[2][24][1 << 12];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  if (n == 1) {
    puts(a[0] == 1 ? "YES" : "NO");
    return 0;
  }
  sort(a, a + n);
  int leaf = 0, inter = 0;
  for (int i = 0; i < n; ++i) {
    if (a[i] == 1) {
      ++leaf;
    } else {
      ++inter;
    }
  }
  if (a[n - 1] != n || inter >= leaf) {
    printf("NO\n");
    return 0;
  }
  memset(dp, 0, sizeof(dp));
  int now = 0;
  dp[now][leaf][0] = true;
  for (int i = 0; i < inter; ++i) {
    int size = a[leaf + i], next = now ^ 1;
    for (int l = 0; l <= leaf; ++l) {
      for (int st = 0; st < (1 << i); ++st) {
        if (!dp[now][l][st]) {
          continue;
        }
        for (int rst = st;; rst = (st) & (rst - 1)) {
          int sum = 1, cnt = 0;
          for (int j = 0; j < i; ++j) {
            if (rst >> j & 1) {
              sum += a[leaf + j];
              ++cnt;
            }
          }
          if (sum <= size && l + sum >= size) {
            int useleaf = size - sum;
            cnt += useleaf;
            if (cnt >= 2) {
              dp[next][l - useleaf][st ^ rst | (1 << i)] = true;
            }
          }
          if (rst == 0) {
            break;
          }
        }
        dp[now][l][st] = false;
      }
    }
    now = next;
  }
  puts(dp[now][0][1 << inter - 1] ? "YES" : "NO");
  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;
const int N = 24;
int n, a[N];
int dp[2][24][1 << 12];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  if (n == 1) {
    puts(a[0] == 1 ? "YES" : "NO");
    return 0;
  }
  sort(a, a + n);
  int leaf = 0, inter = 0;
  for (int i = 0; i < n; ++i) {
    if (a[i] == 1) {
      ++leaf;
    } else {
      ++inter;
    }
  }
  if (a[n - 1] != n || inter >= leaf) {
    printf("NO\n");
    return 0;
  }
  memset(dp, 0, sizeof(dp));
  int now = 0;
  dp[now][leaf][0] = true;
  for (int i = 0; i < inter; ++i) {
    int size = a[leaf + i], next = now ^ 1;
    for (int l = 0; l <= leaf; ++l) {
      for (int st = 0; st < (1 << i); ++st) {
        if (!dp[now][l][st]) {
          continue;
        }
        for (int rst = st;; rst = (st) & (rst - 1)) {
          int sum = 1, cnt = 0;
          for (int j = 0; j < i; ++j) {
            if (rst >> j & 1) {
              sum += a[leaf + j];
              ++cnt;
            }
          }
          if (sum <= size && l + sum >= size) {
            int useleaf = size - sum;
            cnt += useleaf;
            if (cnt >= 2) {
              dp[next][l - useleaf][st ^ rst | (1 << i)] = true;
            }
          }
          if (rst == 0) {
            break;
          }
        }
        dp[now][l][st] = false;
      }
    }
    now = next;
  }
  puts(dp[now][0][1 << inter - 1] ? "YES" : "NO");
  return 0;
}

```