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.
#include <bits/stdc++.h>
using namespace std;
bool visited[220];
int nums[220];
int flow(int ini, int fim, int n, vector<int> *graph, int **cap) {
  int answer = 0;
  int previous[n];
  bool used[n];
  while (true) {
    fill(used, used + n, false);
    fill(previous, previous + n, -1);
    queue<int> fila;
    fila.push(ini);
    while (!fila.empty()) {
      int u = fila.front();
      used[u] = true;
      fila.pop();
      for (int v : graph[u])
        if (!used[v] and cap[u][v] > 0) {
          previous[v] = u;
          fila.push(v);
        }
    }
    if (previous[fim] == -1) break;
    int minimum = INT_MAX, u = fim;
    while (previous[u] != -1) {
      minimum = min(cap[previous[u]][u], minimum);
      u = previous[u];
    }
    answer += minimum;
    u = fim;
    while (previous[u] != -1) {
      cap[previous[u]][u] -= minimum;
      cap[u][previous[u]] += minimum;
      u = previous[u];
    }
  }
  return answer;
}
bool isPrime(int num) {
  if (num % 2 == 0) return false;
  for (int i = 3; i <= sqrt(num); i += 2)
    if (num % i == 0) return false;
  return true;
}
vector<int> dfs(int u, vector<int> graph[], int **cap) {
  vector<int> vec;
  visited[u] = true;
  for (int v : graph[u]) {
    if (!visited[v] && nums[v] % 2 == 0 && cap[u][v] == 0)
      vec = dfs(v, graph, cap);
    else if (!visited[v] && nums[v] % 2 != 0 && cap[u][v] == 1)
      vec = dfs(v, graph, cap);
  }
  vec.push_back(u);
  return vec;
}
int main() {
  int n, x, nume = 0, numo = 0;
  vector<int> connections[220];
  vector<vector<int> > tables;
  int **cap = new int *[220];
  for (int i = 0; i < 220; i++) cap[i] = new int[220];
  for (int i = 0; i < 220; i++)
    for (int j = 0; j < 220; j++) cap[i][j] = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    nums[i] = x;
    if (x % 2 != 0) {
      numo++;
      connections[218].push_back(i);
      connections[i].push_back(218);
      cap[218][i] = 2;
    } else {
      nume++;
      connections[i].push_back(219);
      connections[219].push_back(i);
      cap[i][219] = 2;
    }
  }
  if (n % 2 != 0 || nume != numo)
    printf("Impossible\n");
  else {
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (isPrime(nums[i] + nums[j])) {
          connections[i].push_back(j);
          connections[j].push_back(i);
          if (nums[i] % 2 != 0)
            cap[i][j] = 1;
          else
            cap[j][i] = 1;
        }
      }
    }
    if (flow(218, 219, 220, connections, cap) == n) {
      for (int i = 0; i < n; i++)
        if (!visited[i]) tables.push_back(dfs(i, connections, cap));
      printf("%d\n", tables.size());
      for (int i = 0; i < tables.size(); i++) {
        printf("%d", tables[i].size());
        for (int j = 0; j < tables[i].size(); j++)
          printf(" %d", tables[i][j] + 1);
        printf("\n");
      }
    } else
      printf("Impossible\n");
  }
  return 0;
}
