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 is_prime(int a) {
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) return 0;
  }
  return 1;
}
int n;
bool kuhn(int s, vector<vector<int>> &g, vector<int> &vis, vector<int> &now,
          vector<int> &p) {
  vis[s] = 1;
  for (int to : g[s]) {
    if (now[to - n] == s + n || p[to - n] == s + n) continue;
    if (p[to] == -1) {
      p[to] = s;
      p[s] = to;
      return true;
    }
    if (vis[p[to]]) continue;
    now[s] = to;
    if (kuhn(p[to], g, vis, now, p)) {
      p[to] = s;
      p[s] = to;
      return true;
    }
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<vector<int>> g(n + n);
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) {
      if (is_prime(a[i] + a[j])) {
        g[i].push_back(j + n);
        g[j].push_back(i + n);
      }
    }
  vector<int> p(n + n, -1);
  vector<int> vis(n + n), now(n + n, -1);
  for (int i = 0; i < n; i++) {
    fill(vis.begin(), vis.end(), 0);
    fill(now.begin(), now.end(), -1);
    kuhn(i, g, vis, now, p);
  }
  for (int i = 0; i < p.size(); i++)
    if (p[i] == -1) {
      cout << "Impossible";
      return 0;
    }
  vector<vector<int>> ans;
  fill(vis.begin(), vis.end(), 0);
  for (int i = 0; i < p.size(); i++) {
    if ((i < n && vis[i]) || (i >= n && vis[i - n])) continue;
    int cur = i, q = p[i] - n;
    vis[cur] = 1;
    ans.push_back(vector<int>({cur}));
    while (q != cur) {
      vis[q] = 1;
      ans.back().push_back(q);
      q = p[q] - n;
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].size() << " ";
    for (int j = 0; j < ans[i].size(); j++) cout << ans[i][j] + 1 << " ";
    cout << "\n";
  }
}
