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;
const int N = 205;
vector<int> adj[N], mem[N];
int cap[N][N], edge[N][N], visit[N], p[N], n, source, sink, a[N], cnt, max_flow;
int is_prime(int a) {
  for (int i = 2; i * i <= a; i++)
    if (a % i == 0) return 0;
  return 1;
}
void add(int u, int v, int _cap) {
  adj[u].push_back(v);
  adj[v].push_back(u);
  cap[u][v] = _cap;
  edge[u][v] = 1;
}
int bfs() {
  memset(visit, 0, sizeof visit);
  queue<int> q;
  q.push(source);
  visit[source] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (u == sink) return 1;
    for (int v : adj[u])
      if (cap[u][v] && !visit[v]) {
        visit[v] = 1;
        p[v] = u;
        q.push(v);
      }
  }
  return 0;
}
void dfs(int u) {
  visit[u] = 1;
  mem[cnt].push_back(u);
  for (int v : adj[u])
    if (!visit[v]) dfs(v);
}
int main() {
  cin >> n;
  source = 0;
  sink = n + 1;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] % 2)
      add(source, i, 2);
    else
      add(i, sink, 2);
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i] % 2 && a[j] % 2 == 0 && is_prime(a[i] + a[j])) add(i, j, 1);
  while (bfs()) {
    int flow_path = 1e9;
    for (int v = sink; v != source; v = p[v])
      flow_path = min(flow_path, cap[p[v]][v]);
    max_flow += flow_path;
    for (int v = sink; v != source; v = p[v]) {
      cap[p[v]][v] -= flow_path;
      cap[v][p[v]] += flow_path;
    }
  }
  if (max_flow != n) {
    cout << "Impossible";
    return 0;
  }
  for (int i = 1; i <= n; i++) adj[i].clear();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (edge[i][j] && !cap[i][j]) {
        adj[i].push_back(j);
        adj[j].push_back(i);
      }
  memset(visit, 0, sizeof visit);
  for (int i = 1; i <= n; i++) {
    if (!visit[i]) {
      cnt++;
      dfs(i);
    }
  }
  cout << cnt << '\n';
  for (int i = 1; i <= cnt; i++) {
    cout << mem[i].size() << ' ';
    for (int j : mem[i]) cout << j << ' ';
    cout << '\n';
  }
}