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;
int temp[300][300], graph[300][300];
bool flag[300][300];
bool ar[100000];
bool bfs(int s, int t, int *parent, int v) {
  bool visited[v + 1];
  fill(visited, visited + v + 1, false);
  queue<int> q;
  q.push(s);
  visited[s] = true;
  parent[s] = -1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < v; i++) {
      if (visited[i] == false && temp[u][i] > 0) {
        q.push(i);
        parent[i] = u;
        visited[i] = true;
      }
    }
  }
  return visited[t];
}
int func(int v, int s, int t) {
  int parent[v + 1];
  int flow = 0;
  while (bfs(s, t, parent, v)) {
    int path_flow = INT_MAX;
    for (int i = t; i != s; i = parent[i]) {
      int u = parent[i];
      path_flow = min(path_flow, temp[u][i]);
    }
    for (int i = t; i != s; i = parent[i]) {
      int u = parent[i];
      temp[u][i] = temp[u][i] - path_flow;
      temp[i][u] = temp[i][u] + path_flow;
      graph[u][i] += path_flow;
      graph[i][u] -= path_flow;
    }
    flow = flow + path_flow;
  }
  return flow;
}
void sieve() {
  for (int i = 2; i * i <= 2 * 10000; i++) {
    if (ar[i]) continue;
    for (int j = 2; j * i <= 2 * 10000; j++) ar[j * i] = true;
  }
}
void check(int flow, int *a, int n) {
  int even = 0, odd = 0;
  for (int i = 0; i < n; i++)
    if (a[i + 1] % 2 == 0)
      even++;
    else
      odd++;
  if (!(even == odd && 2 * even == flow)) {
    cout << "Impossible" << endl;
    exit(0);
  }
}
void dfs(int i, list<int> *adj, bool *visited, vector<int> &v) {
  v.push_back(i + 1);
  visited[i] = true;
  list<int>::iterator j;
  for (j = adj[i].begin(); j != adj[i].end(); j++)
    if (!visited[*j]) dfs(*j, adj, visited, v);
}
void fuckthefox(int n, int *a) {
  list<int> adj[n];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j && a[i] % 2 == 1 && a[j] % 2 == 0 && !ar[a[i] + a[j]] &&
          graph[i][j] == 1) {
        adj[i - 1].push_back(j - 1);
        adj[j - 1].push_back(i - 1);
      }
  vector<vector<int> > ans;
  bool visited[n + 2];
  fill(visited, visited + n + 2, false);
  for (int i = 0; i < n; i++)
    if (!visited[i]) {
      vector<int> v;
      dfs(i, adj, visited, v);
      ans.push_back(v);
    }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    vector<int> v = ans[i];
    cout << v.size() << " ";
    for (int j = 0; j < v.size(); j++) cout << v[j] << " ";
    cout << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  sieve();
  int n;
  cin >> n;
  int a[n + 1];
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j && a[i] % 2 == 1 && a[j] % 2 == 0 && !ar[a[i] + a[j]])
        temp[i][j] = 1;
  for (int i = 1; i <= n; i++)
    if (a[i] % 2 == 1)
      temp[0][i] = 2;
    else
      temp[i][n + 1] = 2;
  int flow = func(n + 2, 0, n + 1);
  check(flow, a, n);
  fuckthefox(n, a);
  return 0;
}