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 checked[300];
int n, V[300], T[300], F[300][300], C[300][300];
vector<int> G[300], sol[300];
bool isPrime(int x) {
  for (int d = 2; d <= sqrt(x); d++)
    if (x % d == 0) return false;
  return true;
}
bool BFS() {
  queue<int> Q;
  memset(T, -1, sizeof(T));
  while (!Q.empty()) Q.pop();
  Q.push(0);
  while (!Q.empty()) {
    int node = Q.front();
    Q.pop();
    for (int i = 0; i < G[node].size(); i++) {
      int nextNode = G[node][i];
      if (T[nextNode] == -1 && F[node][nextNode] < C[node][nextNode]) {
        T[nextNode] = node;
        Q.push(nextNode);
        if (nextNode == n + 1) return true;
      }
    }
  }
  return false;
}
void DFS(int node, int nr, int type) {
  checked[node] = true;
  sol[nr].push_back(node);
  for (int i = 0; i < G[node].size(); i++)
    if (checked[G[node][i]] == false && F[node][G[node][i]] == type) {
      DFS(G[node][i], nr, -type);
      return;
    }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> V[i];
    if (V[i] % 2 == 0) {
      G[0].push_back(i);
      C[0][i] = 2;
    } else {
      G[i].push_back(n + 1);
      C[i][n + 1] = 2;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (isPrime(V[i] + V[j])) {
        if (V[i] % 2 == 0) C[i][j] = 1;
        G[i].push_back(j);
        G[j].push_back(i);
      }
  int totalFlow = 0;
  while (BFS()) {
    int node = n + 1, flow = 0x3f3f3f3f;
    while (node != 0) {
      flow = min(flow, C[T[node]][node] - F[T[node]][node]);
      node = T[node];
    }
    node = n + 1;
    while (node != 0) {
      F[T[node]][node] += flow;
      F[node][T[node]] -= flow;
      node = T[node];
    }
    totalFlow += flow;
  }
  if (totalFlow != n) {
    cout << "Impossible";
    return 0;
  }
  int nr = 0;
  for (int i = 1; i <= n; i++)
    if (V[i] % 2 == 0 && !checked[i]) {
      nr++;
      DFS(i, nr, 1);
    }
  cout << nr << "\n";
  for (int i = 1; i <= nr; i++) {
    cout << sol[i].size() << " ";
    for (int j = 0; j < sol[i].size(); j++) cout << sol[i][j] << " ";
    cout << "\n";
  }
}
