Problem: Fox Ciel is participating in a party in Prime Kingdom. There are n foxes there (include Fox Ciel). The i-th fox is ai years old.

They will have dinner around some round tables. You want to distribute foxes such that:

  1. Each fox is sitting at some table. 
  2. Each table has at least 3 foxes sitting around it. 
  3. The sum of ages of any two adjacent foxes around each table should be a prime number. 



If k foxes f1, f2, ..., fk are sitting around table in clockwise order, then for 1 ≤ i ≤ k - 1: fi and fi + 1 are adjacent, and f1 and fk are also adjacent.

If it is possible to distribute the foxes in the desired manner, find out a way to do that.

Input

The first line contains single integer n (3 ≤ n ≤ 200): the number of foxes in this party. 

The second line contains n integers ai (2 ≤ ai ≤ 104).

Output

If it is impossible to do this, output "Impossible".

Otherwise, in the first line output an integer m (<image>): the number of tables.

Then output m lines, each line should start with an integer k -=– the number of foxes around that table, and then k numbers — indices of fox sitting around that table in clockwise order.

If there are several possible arrangements, output any of them.

Examples

Input

4
3 4 8 9


Output

1
4 1 2 4 3


Input

5
2 2 2 2 2


Output

Impossible


Input

12
2 3 4 5 6 7 8 9 10 11 12 13


Output

1
12 1 2 3 6 5 12 9 8 7 10 11 4


Input

24
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25


Output

3
6 1 2 3 6 5 4
10 7 8 9 12 15 14 13 16 11 10
8 17 18 23 22 19 20 21 24

Note

In example 1, they can sit around one table, their ages are: 3-8-9-4, adjacent sums are: 11, 17, 13 and 7, all those integers are primes.

In example 2, it is not possible: the sum of 2+2 = 4 is not a prime number.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int prime[100001];
vector<int> kraw[201];
vector<int> cykle[201];
int tab[201];
int sko[201];
int odw[201];
int ile = 0;
int czy[201];
int mach(int x) {
  odw[x] = ile;
  for (vector<int>::iterator it = kraw[x].begin(); it != kraw[x].end(); it++) {
    if (sko[*it] == 0 && sko[x] != *it && odw[*it] != ile) {
      sko[*it] = x;
      return 1;
    }
  }
  for (vector<int>::iterator it = kraw[x].begin(); it != kraw[x].end(); it++) {
    if (odw[sko[*it]] != ile && sko[x] != *it && mach(sko[*it]) == 1) {
      sko[*it] = x;
      return 1;
    }
  }
  return 0;
}
int main() {
  int n;
  int wyn = 0;
  int m = 100000;
  prime[1] = 1;
  for (int i = 2; i <= m; i++) {
    if (prime[i] == 0)
      for (int j = 2 * i; j <= m; j += i) prime[j] = 1;
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &tab[i]);
  }
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      if (prime[tab[i] + tab[j]] == 0) {
        kraw[i].push_back(j);
        kraw[j].push_back(i);
      }
    }
  for (int i = 1; i <= n; i++) {
    ile++;
    wyn += mach(i);
  }
  if (wyn < n)
    printf("Impossible");
  else {
    ile = 0;
    for (int i = 1; i <= n; i++) {
      if (czy[i] == 0) {
        ile++;
        int gdzie = i;
        while (czy[gdzie] == 0) {
          cykle[ile].push_back(gdzie);
          czy[gdzie] = 1;
          gdzie = sko[gdzie];
        }
      }
    }
    printf("%d\n", ile);
    for (int i = 1; i <= ile; i++) {
      printf("%d ", cykle[i].size());
      for (int j = 0; j < cykle[i].size(); j++) printf("%d ", cykle[i][j]);
      printf("\n");
    }
  }
}