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;
int a[200];
bool can_go[200][200];
vector<int> g[200];
bool left_used[200];
int right_used[200];
bool in_matching[200][200];
bool is_prime(int x) {
  for (int y(2); y * y <= x; ++y) {
    if (x % y == 0) return false;
  }
  return true;
}
bool Kuhn(int v) {
  if (left_used[v]) return false;
  left_used[v] = true;
  for (int u : g[v]) {
    if (!can_go[v][u]) continue;
    if (-1 == right_used[u]) {
      can_go[u][v] = false;
      right_used[u] = v;
      return true;
    }
    if (Kuhn(right_used[u])) {
      can_go[u][right_used[u]] = true;
      can_go[u][v] = false;
      right_used[u] = v;
      return true;
    }
  }
  return false;
}
int p[200];
int find_set(int v) { return p[v] == v ? v : p[v] = find_set(p[v]); }
void unite_set(int u, int v) {
  u = find_set(u);
  v = find_set(v);
  if (u != v) p[v] = u;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i(0); i < n; ++i) cin >> a[i];
  for (int i(0); i < n; ++i) {
    for (int j{i + 1}; j < n; ++j) {
      if (is_prime(a[i] + a[j])) {
        can_go[i][j] = can_go[j][i] = true;
        g[i].push_back(j);
        g[j].push_back(i);
      }
    }
  }
  int cnt(0);
  memset(right_used, -1, sizeof right_used);
  for (int i(0); i < n; ++i) {
    if (Kuhn(i)) {
      memset(left_used, false, sizeof left_used);
      ++cnt;
    }
  }
  if (cnt < n) {
    cout << "Impossible\n";
    return 0;
  }
  for (int i(0); i < n; ++i) p[i] = i;
  for (int i(0); i < n; ++i) unite_set(i, right_used[i]);
  map<int, vector<int>> tables;
  for (int i(0); i < n; ++i) tables[find_set(i)].push_back(i);
  for (const auto& t : tables)
    if (t.second.size() < 3) {
      cout << "Impossible\n";
      return 0;
    }
  cout << tables.size() << '\n';
  memset(left_used, false, sizeof left_used);
  for (int i(0); i < n; ++i) {
    if (left_used[i]) continue;
    vector<int> cur;
    for (int j(i); !left_used[j]; j = right_used[j]) {
      left_used[j] = true;
      cur.push_back(j);
    }
    cout << cur.size();
    for (int t : cur) cout << ' ' << 1 + t;
    cout << '\n';
  }
}
