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 = 200 + 121;
const int M = 100 * 1000 + 121;
const int E = 2 * 200 * 200 + 121;
int n, INF = 1e9 + 121, a[N], cap[E], from[E], to[E], prv[E], head[N],
       source = N - 2, sink = N - 1, ecnt, ans;
vector<int> adj[N], vans[N];
bool isprime[M];
bool mark[N];
void add_edge(int u, int v, int uv, int vu = 0) {
  from[ecnt] = u;
  to[ecnt] = v;
  cap[ecnt] = uv;
  prv[ecnt] = head[u];
  head[u] = ecnt++;
  from[ecnt] = v;
  to[ecnt] = u;
  cap[ecnt] = vu;
  prv[ecnt] = head[v];
  head[v] = ecnt++;
}
int dfs(int v, int flow = INF) {
  if (v == sink) return flow;
  if (flow == 0 || mark[v]) return 0;
  mark[v] = 1;
  for (int i = head[v]; i != -1; i = prv[i]) {
    int x = dfs(to[i], min(flow, cap[i]));
    if (x > 0) {
      cap[i] -= x;
      cap[i ^ 1] += x;
      return x;
    }
  }
  return 0;
}
int max_flow() {
  int ans = 0, f;
  memset(mark, false, sizeof mark);
  while (f = dfs(source)) {
    ans += f;
    memset(mark, false, sizeof mark);
  }
  return ans;
}
void dfs2(int w, int v) {
  vans[w].push_back(v);
  mark[v] = true;
  for (auto u : adj[v]) {
    if (mark[u]) continue;
    dfs2(w, u);
  }
}
int main() {
  cin >> n;
  fill(isprime + 2, isprime + M, 1);
  memset(head, -1, sizeof head);
  for (int i = 2; i < M; i++)
    if (isprime[i])
      for (int j = i + i; j < M; j += i) isprime[j] = 0;
  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 (isprime[a[i] + a[j]]) {
        if (a[i] & 1)
          add_edge(i, j, 1);
        else
          add_edge(j, i, 1);
      }
  for (int i = 0; i < n; i++)
    if (a[i] & 1)
      add_edge(source, i, 2);
    else
      add_edge(i, sink, 2);
  int t = max_flow();
  if (t != n) {
    cout << "Impossible"
         << "\n";
    return 0;
  }
  for (int i = 0; i < ecnt; i++)
    if (cap[i] == 0 && (a[from[i]] & 1) && from[i] < n && to[i] < n) {
      int u = from[i], v = to[i];
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
  memset(mark, false, sizeof mark);
  for (int i = 0; i < n; i++) {
    if (mark[i]) continue;
    dfs2(ans++, i);
  }
  cout << ans << "\n";
  for (int i = 0; i < ans; i++) {
    cout << vans[i].size() << " ";
    for (auto j : vans[i]) cout << j + 1 << " ";
    cout << "\n";
  }
  return 0;
}