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>
int a[200];
int d[202][202];
int v[202];
int r[202][202];
int s[202][202];
int sn[202];
int snn;
int f(int x) {
  int i;
  for (i = 2; i * i <= x; i++)
    if (x % i == 0) return 0;
  return 1;
}
int find(int x) {
  if (x == 201) return 1;
  if (v[x]) return 0;
  int i;
  v[x] = 1;
  for (i = 1; i <= 201; i++)
    if (d[x][i] && find(i)) {
      d[x][i]--;
      d[i][x]++;
      return 1;
    }
  return 0;
}
void dfs(int x) {
  if (v[x]) return;
  int i;
  v[x] = 1;
  s[snn][sn[snn]++] = x;
  for (i = 1; i <= 200; i++)
    if (r[x][i]) dfs(i);
}
int main() {
  int i, j, n;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] % 2)
      d[0][i] = 2;
    else
      d[i][201] = 2;
  }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++)
      if (a[i] % 2 && a[j] % 2 == 0 && f(a[i] + a[j])) d[i][j] = 1;
  for (i = 0; find(0); i++)
    for (j = 0; j <= 201; j++) v[j] = 0;
  if (i < n) {
    puts("Impossible");
    return 0;
  }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++)
      if (a[i] % 2 && a[j] % 2 == 0 && f(a[i] + a[j]) && !d[i][j])
        r[i][j] = r[j][i] = 1;
  for (i = 1; i <= n; i++) v[i] = 0;
  for (i = 1; i <= n; i++)
    if (!v[i]) {
      dfs(i);
      snn++;
    }
  printf("%d\n", snn);
  for (i = 0; i < snn; i++) {
    printf("%d ", sn[i]);
    for (j = 0; j < sn[i]; j++) printf("%d ", s[i][j]);
    puts("");
  }
}