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 N, M;
int arr[201];
int capacity[1001][1001];
bool visited[1001];
int ans = 0;
bool used[201];
vector<int> v[201];
int cur = 0;
void dfs2(int node) {
  used[node] = true;
  v[cur].push_back(node);
  for (int i = 1; i <= N; i++) {
    if (!used[i]) {
      if (arr[node] % 2 == 1 && capacity[i][node] == 1)
        dfs2(i);
      else if (arr[node] % 2 == 0 && capacity[node][i] == 1)
        dfs2(i);
    }
  }
}
bool isprime(int X) {
  for (int i = 2; i * i <= X; i++) {
    if (X % i == 0) return false;
  }
  return true;
}
int dfs(int node, int curmin) {
  if (node == 0)
    for (int i = 0; i <= N + 1; i++) visited[i] = false;
  visited[node] = true;
  if (node == N + 1) {
    ans += curmin;
    return curmin;
  }
  for (int i = 0; i <= N + 1; i++) {
    if (capacity[node][i] != 0 && !visited[i]) {
      int temp = dfs(i, min(curmin, capacity[node][i]));
      capacity[node][i] -= temp;
      capacity[i][node] += temp;
      if (temp != 0) return temp;
    }
  }
  return 0;
}
int main() {
  int even = 0;
  cin >> N;
  if (N % 2 == 1) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (int i = 1; i <= N; i++) cin >> arr[i];
  for (int i = 1; i <= N; i++) {
    if (arr[i] % 2 == 1)
      capacity[0][i] = 2;
    else {
      even++;
      capacity[i][N + 1] = 2;
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) {
      if (i != j && isprime(arr[i] + arr[j])) {
        if (arr[i] % 2 == 1)
          capacity[i][j] = 1;
        else
          capacity[j][i] = 1;
      }
    }
  }
  while (dfs(0, 1001001001))
    ;
  if (ans != even * 2) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (int i = 1; i <= N; i++) {
    if (!used[i]) {
      dfs2(i);
      cur++;
    }
  }
  cout << cur << endl;
  for (int i = 0; i < cur; i++) {
    cout << v[i].size() << ' ';
    for (int j = 0; j < v[i].size(); j++) cout << v[i][j] << ' ';
    cout << endl;
  }
}