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;
int A[205], P[405], N, C[405][405], T, U[20005], V[405], s, t = 201, f, n;
queue<int> Q;
vector<int> E[405], E2[405], B[405];
void dfs(int n, int r) {
  if (V[n]) return;
  B[T].push_back(n);
  V[n] = 1;
  for (int v : E2[n]) dfs(v, r);
}
int main() {
  for (int i = 2; i <= (20000); i++)
    if (!U[i])
      for (n = 2 * i; n <= 20000; n += i) U[n] = 1;
  cin >> N;
  for (int i = 1; i <= (N); i++) cin >> A[i];
  for (int i = 1; i <= (N); i++)
    for (int j = 1; j <= (N); j++)
      if (A[i] % 2 == 0 && A[j] % 2 == 1 && !U[A[i] + A[j]])
        E[i].push_back(j), E[j].push_back(i), C[i][j] = 1;
  for (int i = 1; i <= (N); i++) {
    if (A[i] % 2 == 0)
      E[s].push_back(i), C[s][i] = 2;
    else
      E[i].push_back(t), C[i][t] = 2;
  }
  while (1) {
    for (int i = 1; i <= (201); i++) V[i] = 0;
    Q.push(0);
    V[0] = 1;
    while (!Q.empty()) {
      int n = Q.front();
      Q.pop();
      for (int v : E[n])
        if (C[n][v] && !V[v]) V[v] = 1, P[v] = n, Q.push(v);
    }
    if (!V[t]) break;
    f++, n = t;
    while (n) {
      C[P[n]][n]--;
      C[n][P[n]]++;
      n = P[n];
    }
  }
  if (f < N) {
    cout << "Impossible";
    return 0;
  }
  for (int i = 1; i <= (201); i++) V[i] = 0;
  for (int i = 1; i <= (N); i++)
    if (A[i] % 2 == 0)
      for (int v : E[i])
        if (!C[i][v]) E2[i].push_back(v), E2[v].push_back(i);
  for (int i = 1; i <= (N); i++)
    if (!V[i]) T++, dfs(i, i);
  cout << T << "\n";
  for (int i = 1; i <= (T); i++) {
    cout << B[i].size() << " ";
    for (int k : B[i]) cout << k << " ";
    cout << "\n";
  }
}
