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;
vector<int> v[210], w[210], sol[210];
queue<int> cd;
int nrsol;
int C[210][210];
int F[210][210];
bool viz[210], viz2[210];
int ant[210];
int S, D;
int A[210];
bool BFS(int s) {
  int i;
  for (i = S; i <= D; ++i) viz[i] = false;
  viz[s] = true;
  cd.push(s);
  while (!cd.empty()) {
    s = cd.front();
    cd.pop();
    if (s != D)
      for (i = 0; i < v[s].size(); ++i)
        if (!viz[v[s][i]] && F[s][v[s][i]] < C[s][v[s][i]]) {
          ant[v[s][i]] = s;
          viz[v[s][i]] = true;
          cd.push(v[s][i]);
        }
  }
  return viz[D];
}
void DFS(int x) {
  viz2[x] = true;
  for (int i = 0; i < w[x].size(); ++i)
    if (!viz2[w[x][i]]) DFS(w[x][i]);
  sol[nrsol].push_back(x);
}
int main() {
  int i, j, k;
  cin >> n;
  for (i = 1; i <= n; ++i) cin >> A[i];
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) {
      int val = A[i] + A[j];
      bool prim = true;
      for (k = 2; k * k <= val; ++k)
        if (val % k == 0) prim = false;
      if (prim && A[i] % 2 == 0) {
        v[i].push_back(j);
        v[j].push_back(i);
        C[i][j] = 1;
      }
    }
  S = 0;
  D = n + 1;
  for (i = 1; i <= n; ++i)
    if (A[i] % 2 == 0) {
      v[S].push_back(i);
      v[i].push_back(S);
      C[S][i] = 2;
    } else {
      v[D].push_back(i);
      v[i].push_back(D);
      C[i][D] = 2;
    }
  int flow = 0, newflow, s;
  while (BFS(S))
    for (i = 0; i < v[D].size(); ++i) {
      ant[D] = v[D][i];
      newflow = 2100000000;
      s = D;
      while (s != S) {
        newflow = min(newflow, C[ant[s]][s] - F[ant[s]][s]);
        s = ant[s];
      }
      s = D;
      while (s != S) {
        F[ant[s]][s] += newflow;
        F[s][ant[s]] -= newflow;
        s = ant[s];
      }
      flow += newflow;
    }
  if (flow != n)
    cout << "Impossible\n";
  else {
    for (i = 1; i <= n; ++i)
      for (j = 1; j <= n; ++j)
        if (F[i][j] == 1) {
          w[i].push_back(j);
          w[j].push_back(i);
        }
    for (i = 1; i <= n; ++i)
      if (!viz2[i]) {
        nrsol++;
        DFS(i);
      }
    cout << nrsol << "\n";
    for (i = 1; i <= nrsol; ++i) {
      cout << sol[i].size() << " ";
      for (j = 0; j < sol[i].size(); ++j) cout << sol[i][j] << " ";
      cout << "\n";
    }
  }
  return 0;
}