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;
long long G(long long a, long long b) {
  if (b == 0) return a;
  return G(b, a % b);
}
long long a[222], adj[222][222], par[222], con[222][222], n;
bool mark[222];
queue<long long> q;
vector<vector<long long> > ans;
vector<long long> v;
long long bfs(long long source, long long sink) {
  memset(mark, 0, sizeof(mark));
  mark[source] = 1;
  q.push(source);
  while (!q.empty()) {
    long long at = q.front();
    q.pop();
    for (int i = 0; i < n + 2; i++) {
      if (!mark[i] && adj[at][i] > 0) {
        mark[i] = 1;
        q.push(i);
        par[i] = at;
      }
    }
  }
  if (!mark[sink]) return 0;
  long long tmp = sink;
  while (tmp != source) {
    adj[par[tmp]][tmp]--;
    adj[tmp][par[tmp]]++;
    if (tmp != sink && par[tmp] != source) {
      if (a[par[tmp]] & 1) {
        con[tmp][par[tmp]]++;
        con[par[tmp]][tmp]++;
      } else {
        con[tmp][par[tmp]]--;
        con[par[tmp]][tmp]--;
      }
    }
    tmp = par[tmp];
  }
  return 1;
}
long long find(long long source, long long sink) {
  long long res = 0, tmp = 0;
  while (tmp = bfs(source, sink) != 0) res += tmp;
  return res;
}
void dfs(long long r) {
  mark[r] = 1;
  v.push_back(r);
  for (int i = 0; i < n; i++) {
    if (!mark[i] && con[r][i] > 0) dfs(i);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  long long o = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] & 1) {
      o++;
      adj[n][i] = 2;
    } else
      adj[i][n + 1] = 2;
  }
  for (int i = 0; i < n; i++) {
    if ((a[i] % 2) == 0) continue;
    for (int j = 0; j < n; j++) {
      bool flag = 1;
      long long tmp = a[i] + a[j];
      for (int k = 2; k * k <= tmp; k++) {
        if (tmp % k == 0) flag = 0;
      }
      if (flag) adj[i][j] = 1;
    }
  }
  if (2 * o != n || find(n, n + 1) != n) {
    cout << "Impossible\n";
    return 0;
  }
  memset(mark, 0, sizeof(mark));
  for (int i = 0; i < n; i++) {
    if (!mark[i]) {
      v.clear();
      dfs(i);
      ans.push_back(v);
    }
  }
  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 << endl;
  }
}