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 nmax = 2 * 200 + 42;
int n;
int a[nmax];
vector<int> adj[nmax];
bool prime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return 0;
  return 1;
}
int parent[nmax];
int flow[nmax][nmax];
void dfs(int node, int par) {
  if (parent[node] != -1) return;
  parent[node] = par;
  for (auto k : adj[node])
    if (flow[node][k]) dfs(k, node);
}
set<int> matched[nmax];
vector<int> path[nmax];
bool been[nmax];
int pos = 0;
void make_path(int node) {
  if (been[node]) return;
  been[node] = 1;
  path[pos].push_back(node);
  for (auto k : adj[node]) make_path(k);
}
int main() {
  cin >> n;
  int even = 0, odd = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] % 2 == 0)
      even++;
    else
      odd++;
  }
  if (even != odd) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i] % 2 == 1 && a[j] % 2 == 0 && prime(a[i] + a[j])) {
        adj[i].push_back(j + n);
        adj[j + n].push_back(i);
        flow[i][j + n] = 1;
      }
  for (int i = 1; i <= n; i++)
    if (a[i] % 2 == 1) {
      adj[0].push_back(i);
      flow[0][i] = 2;
    }
  for (int j = 1; j <= n; j++)
    if (a[j] % 2 == 0) {
      adj[n + j].push_back(2 * n + 1);
      flow[n + j][2 * n + 1] = 2;
    }
  int f = 0;
  while (1) {
    memset(parent, -1, sizeof(parent));
    dfs(0, 0);
    if (parent[2 * n + 1] == -1) break;
    f++;
    int node = 2 * n + 1, t = 0;
    while (node) {
      if (node != 2 * n + 1 && parent[node] != 0) {
        if (t % 2 == 0) {
          matched[node].erase(parent[node]);
        } else {
          matched[parent[node]].insert(node);
        }
      }
      t++;
      flow[parent[node]][node]--;
      flow[node][parent[node]]++;
      node = parent[node];
    }
  }
  if (f != n) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (int i = 0; i <= 2 * n + 1; i++) adj[i].clear();
  for (int i = 1; i <= n; i++)
    if (a[i] % 2 == 1) {
      for (auto k : matched[i]) {
        adj[i].push_back(k - n);
        adj[k - n].push_back(i);
      }
    }
  memset(been, 0, sizeof(been));
  for (int i = 1; i <= n; i++)
    if (been[i] == 0) {
      pos++;
      make_path(i);
    }
  cout << pos << endl;
  for (int i = 1; i <= pos; i++) {
    cout << path[i].size();
    for (auto k : path[i]) cout << " " << k;
    cout << endl;
  }
  return 0;
}
