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[1 << 17];
int u[1 << 17];
int n;
int a[222];
int C[222][222];
int F[222][222];
int used[222];
int dfs(int x, int y) {
  if (used[x]) return 0;
  used[x] = 1;
  if (x == y) return 1;
  for (int i = 0; i < n + 2; ++i)
    if (C[x][i] - F[x][i] > 0 && dfs(i, y)) {
      F[x][i]++;
      F[i][x]--;
      return 1;
    }
  return 0;
}
void go(int x, vector<int>& cycle) {
  if (used[x]) return;
  used[x] = 1;
  cycle.push_back(x + 1);
  for (int i = 0; i < n; ++i)
    if (F[x][i]) go(i, cycle);
}
int main() {
  for (int i = 2; i < 1 << 17; ++i) {
    if (!u[i]) {
      prime[i] = 1;
      for (int j = i; j < 1 << 17; j += i) u[j] = 1;
    }
  }
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      if (i == j || (!(a[i] & 1))) continue;
      if (prime[a[i] + a[j]]) C[i][j] = 1;
    }
  for (int i = 0; i < n; ++i) {
    if (a[i] & 1)
      C[n][i] = 2;
    else
      C[i][n + 1] = 2;
  }
  int res = 0;
  while (1) {
    memset(used, 0, sizeof(used));
    if (!dfs(n, n + 1)) break;
    res++;
  }
  if (res != n)
    printf("Impossible\n");
  else {
    memset(used, 0, sizeof(used));
    vector<vector<int> > res;
    for (int i = 0; i < n; ++i) {
      if (!used[i]) {
        vector<int> cycle;
        go(i, cycle);
        res.push_back(cycle);
      }
    }
    printf("%d\n", res.size());
    for (int i = 0; i < res.size(); ++i) {
      printf("%d ", res[i].size());
      for (int j = 0; j < res[i].size(); ++j) printf("%d ", res[i][j]);
      printf("\n");
    }
  }
  return 0;
}