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;
const int N = 405;
bool vis[N], p[20002];
int match[N], n, a[N];
vector<int> adj[N];
bool dfs(int x) {
  if (x == -1) return 0;
  int i, u;
  if (vis[x]) return 0;
  vis[x] = 1;
  for (i = 0; i < adj[x].size(); ++i) {
    u = adj[x][i];
    if (match[n + x] != u - n && (match[u] == -1 || dfs(match[u]))) {
      match[u] = x;
      return 1;
    }
  }
  return 0;
}
int main() {
  int u, i, j;
  for (i = 2; i < 20001; ++i)
    if (!p[i])
      for (j = i + i; j < 20001; j += i) p[j] = 1;
  cin >> n;
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j)
      if (!p[a[i] + a[j]]) adj[i].push_back(n + j);
  memset(match, -1, sizeof(match));
  for (i = 1; i <= n; ++i) {
    memset(vis, 0, sizeof(vis));
    if (!dfs(i)) {
      printf("Impossible\n");
      return 0;
    }
  }
  memset(vis, 0, sizeof(vis));
  vector<vector<int> > ret;
  vector<int> R;
  for (i = 1; i <= n; ++i) {
    if (!vis[i]) {
      R.clear();
      R.push_back(i);
      u = match[i + n];
      vis[i] = 1;
      while (u != i) {
        R.push_back(u);
        vis[u] = 1;
        u = match[u + n];
      }
      ret.push_back(R);
    }
  }
  cout << ret.size() << endl;
  for (i = 0; i < ret.size(); ++i) {
    cout << ret[i].size() << " ";
    for (j = 0; j < ret[i].size(); ++j) printf("%d ", ret[i][j]);
    cout << endl;
  }
}
