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;
bool dfs(int at, int t, vector<vector<int> > &c, vector<vector<int> > &f,
         vector<int> &mark) {
  if (at == t) return true;
  if (mark[at]) return false;
  mark[at] = true;
  for (int i = 0; i < (int)c.size(); ++i) {
    if (f[at][i] < c[at][i] && dfs(i, t, c, f, mark)) {
      ++f[at][i];
      --f[i][at];
      return true;
    }
  }
  return false;
}
int flow(int n, vector<vector<int> > &c, vector<vector<int> > &f, int s,
         int t) {
  vector<int> mark(n, 0);
  int ans = 0;
  while (dfs(s, t, c, f, mark)) {
    ++ans;
    fill((mark).begin(), (mark).end(), 0);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int &x : a) cin >> x;
  vector<int> primes(2e4 + 1, 1);
  for (int i = 2; i < primes.size(); ++i) {
    for (int j = i * i; j < primes.size(); j += i) primes[j] = 0;
  }
  vector<vector<int> > c(n + 2, vector<int>(n + 2)),
      f(n + 2, vector<int>(n + 2));
  int s = n;
  int t = s + 1;
  for (int i = 0; i < n; ++i) {
    if (a[i] % 2 == 0) {
      c[s][i] = 2;
    } else {
      c[i][t] = 2;
    }
    for (int j = 0; j < n; ++j)
      if (primes[a[i] + a[j]]) {
        if (a[i] % 2 == 0)
          c[i][j] = 1;
        else
          c[j][i] = 1;
      }
  }
  if (flow(n + 2, c, f, s, t) != n) {
    cout << "Impossible\n";
  } else {
    vector<vector<int> > cycles;
    vector<int> mark(n);
    for (int i = 0; i < n; ++i)
      if (!mark[i]) {
        vector<int> cycle;
        int cur = i;
        while (true) {
          mark[cur] = true;
          cycle.push_back(cur + 1);
          bool found = false;
          for (int j = 0; j < n; ++j)
            if (f[cur][j] != 0 && !mark[j]) {
              cur = j;
              found = true;
              break;
            }
          if (!found) break;
        }
        cycles.push_back(cycle);
      }
    cout << cycles.size() << endl;
    for (vector<int> &cycle : cycles) {
      cout << cycle.size();
      for (int v : cycle) cout << " " << v;
      cout << endl;
    }
  }
  return 0;
}