#include <bits/stdc++.h>
using namespace std;
int dp[1 << 11][24];
int main() {
  int n, c = 0, i, j, k, l;
  vector<int> v;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    v.push_back(x);
  }
  sort(v.begin(), v.end());
  for (i = 0; i < n; i++) {
    if (v[i] == 1) c++;
    if (v[i] == 2) {
      puts("NO");
      return 0;
    }
  }
  if (n - c > 11 || v[n - 1] != n) {
    puts("NO");
    return 0;
  }
  if (n == 1) {
    puts("YES");
    return 0;
  }
  dp[0][c] = 1;
  for (i = c; i < n; i++) {
    int x = i - c;
    for (j = 0; j < (1 << x); j++) {
      for (k = 0; k <= c; k++) {
        int y = j;
        if (dp[j][k] == 0) continue;
        do {
          int sum = 1;
          for (l = 0; l < x; l++) {
            if ((y >> l) & 1) {
              if (v[l + c] == v[i] - 1) break;
              sum += v[l + c];
            }
          }
          if (l == x && sum <= v[i] && sum + k >= v[i])
            dp[(j ^ y) | (1 << x)][k - v[i] + sum] = 1;
          y = (y - 1) & j;
        } while (y != j);
      }
    }
  }
  if (dp[1 << (n - c - 1)][0] == 1) {
    puts("YES");
  } else {
    puts("NO");
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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;
int dp[1 << 11][24];
int main() {
  int n, c = 0, i, j, k, l;
  vector<int> v;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    v.push_back(x);
  }
  sort(v.begin(), v.end());
  for (i = 0; i < n; i++) {
    if (v[i] == 1) c++;
    if (v[i] == 2) {
      puts("NO");
      return 0;
    }
  }
  if (n - c > 11 || v[n - 1] != n) {
    puts("NO");
    return 0;
  }
  if (n == 1) {
    puts("YES");
    return 0;
  }
  dp[0][c] = 1;
  for (i = c; i < n; i++) {
    int x = i - c;
    for (j = 0; j < (1 << x); j++) {
      for (k = 0; k <= c; k++) {
        int y = j;
        if (dp[j][k] == 0) continue;
        do {
          int sum = 1;
          for (l = 0; l < x; l++) {
            if ((y >> l) & 1) {
              if (v[l + c] == v[i] - 1) break;
              sum += v[l + c];
            }
          }
          if (l == x && sum <= v[i] && sum + k >= v[i])
            dp[(j ^ y) | (1 << x)][k - v[i] + sum] = 1;
          y = (y - 1) & j;
        } while (y != j);
      }
    }
  }
  if (dp[1 << (n - c - 1)][0] == 1) {
    puts("YES");
  } else {
    puts("NO");
  }
  return 0;
}

```