#include <bits/stdc++.h>
using namespace std;
int n;
int c[30];
int s[1 << 13];
char dp[13][1 << 13][25];
char f(int pos, int mask, int leaves) {
  if (c[pos] == 1) {
    return leaves == 0 ? 1 : 0;
  }
  if ((mask >> pos) & 1) {
    return 0;
  }
  char &res = dp[pos][mask][leaves];
  if (res != -1) {
    return res;
  }
  res = 0;
  for (int nmask = mask;; nmask = (nmask - 1) & mask) {
    int tmp = __builtin_popcount(nmask);
    if (s[nmask] <= c[pos] - 1) {
      int take = c[pos] - 1 - s[nmask];
      if (take <= leaves && take + tmp > 1) {
        char cres = f(pos + 1, mask ^ nmask, leaves - take);
        res |= cres;
      }
    }
    if (nmask == 0) {
      break;
    }
  }
  42;
  ;
  return res;
}
int main() {
  scanf("%d", &n);
  int leaves = 0;
  bool bad = false;
  for (int i = 0; i < n; i++) {
    scanf("%d", &c[i]);
    leaves += c[i] == 1;
    bad |= c[i] == 2;
  }
  if (n == 1) {
    printf("%s\n", c[0] == 1 ? "YES" : "NO");
    return 0;
  }
  int inner = n - leaves;
  if (2 * leaves <= n || bad) {
    puts("NO");
  } else {
    sort(c + 0, c + n);
    reverse(c + 0, c + n);
    memset(dp, -1, sizeof(dp));
    42;
    ;
    for (int mask = 0; mask < (1 << inner); mask++) {
      s[mask] = 0;
      for (int i = 0; i < inner; i++) {
        if ((mask >> i) & 1) {
          s[mask] += c[i];
        }
      }
    }
    printf("%s\n", f(0, ((1 << inner) - 1) ^ 1, leaves) == 1 ? "YES" : "NO");
  }
  return 0;
}

### 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;
int n;
int c[30];
int s[1 << 13];
char dp[13][1 << 13][25];
char f(int pos, int mask, int leaves) {
  if (c[pos] == 1) {
    return leaves == 0 ? 1 : 0;
  }
  if ((mask >> pos) & 1) {
    return 0;
  }
  char &res = dp[pos][mask][leaves];
  if (res != -1) {
    return res;
  }
  res = 0;
  for (int nmask = mask;; nmask = (nmask - 1) & mask) {
    int tmp = __builtin_popcount(nmask);
    if (s[nmask] <= c[pos] - 1) {
      int take = c[pos] - 1 - s[nmask];
      if (take <= leaves && take + tmp > 1) {
        char cres = f(pos + 1, mask ^ nmask, leaves - take);
        res |= cres;
      }
    }
    if (nmask == 0) {
      break;
    }
  }
  42;
  ;
  return res;
}
int main() {
  scanf("%d", &n);
  int leaves = 0;
  bool bad = false;
  for (int i = 0; i < n; i++) {
    scanf("%d", &c[i]);
    leaves += c[i] == 1;
    bad |= c[i] == 2;
  }
  if (n == 1) {
    printf("%s\n", c[0] == 1 ? "YES" : "NO");
    return 0;
  }
  int inner = n - leaves;
  if (2 * leaves <= n || bad) {
    puts("NO");
  } else {
    sort(c + 0, c + n);
    reverse(c + 0, c + n);
    memset(dp, -1, sizeof(dp));
    42;
    ;
    for (int mask = 0; mask < (1 << inner); mask++) {
      s[mask] = 0;
      for (int i = 0; i < inner; i++) {
        if ((mask >> i) & 1) {
          s[mask] += c[i];
        }
      }
    }
    printf("%s\n", f(0, ((1 << inner) - 1) ^ 1, leaves) == 1 ? "YES" : "NO");
  }
  return 0;
}

```