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;
bool P[20100], v[210];
int F[210][210], n, w[210], Flow;
int Res[210][210], cnt, C[210];
int DFS(int a, int ff) {
  if (a == n + 1) {
    Flow += ff;
    return ff;
  }
  int i, t;
  v[a] = true;
  for (i = 0; i <= n + 1; i++) {
    if (!v[i] && F[a][i]) {
      t = DFS(i, min(ff, F[a][i]));
      if (!t) continue;
      F[a][i] -= t;
      F[i][a] += t;
      return t;
    }
  }
  return 0;
}
void Do(int a) {
  int i;
  while (1) {
    v[a] = true;
    Res[cnt][++C[cnt]] = a;
    for (i = 1; i <= n; i++) {
      if (v[i]) continue;
      if (w[a] % 2 == 0 && w[i] % 2 == 1 && F[i][a]) break;
      if (w[a] % 2 == 1 && w[i] % 2 == 0 && F[a][i]) break;
    }
    if (i == n + 1) break;
    a = i;
  }
}
int main() {
  int i, c = 0, j;
  for (i = 2; i <= 20000; i++) {
    if (!P[i]) {
      for (j = i + i; j <= 20000; j += i) P[j] = true;
    }
  }
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    if (w[i] % 2 == 0) {
      c++;
      F[0][i] = 2;
    } else {
      F[i][n + 1] = 2;
    }
  }
  if (c * 2 != n) {
    printf("Impossible\n");
    return 0;
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (w[i] % 2 == 0 && w[j] % 2 == 1 && !P[w[i] + w[j]]) F[i][j] = 1;
    }
  }
  while (1) {
    for (i = 0; i <= n + 1; i++) v[i] = false;
    if (!DFS(0, 2)) break;
  }
  if (Flow != n) {
    printf("Impossible\n");
    return 0;
  }
  for (i = 1; i <= n; i++) v[i] = false;
  for (i = 1; i <= n; i++) {
    if (!v[i]) {
      ++cnt;
      Do(i);
    }
  }
  printf("%d\n", cnt);
  for (i = 1; i <= cnt; i++) {
    printf("%d ", C[i]);
    for (j = 1; j <= C[i]; j++) printf("%d ", Res[i][j]);
    printf("\n");
  }
}