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 N = 2e2 + 7;
const int E = 1e6 + 7;
const int I = 1e9 + 7;
void shit() {
  cout << "Impossible\n";
  exit(0);
}
int a[N];
int prv[E], head[N], to[E], from[E], cap[E];
int ecur;
bool mark[N];
vector<int> adj[N];
void edge_magic(int u, int v, int uv) {
  cap[ecur] = uv;
  prv[ecur] = head[u];
  head[u] = ecur;
  from[ecur] = u, to[ecur] = v;
  ecur++;
}
void add_edge(int u, int v, int uv, int vu = 0) {
  edge_magic(u, v, uv);
  edge_magic(v, u, vu);
}
bool is_prime(int x) {
  for (int i = 2; i * i <= x; ++i)
    if (x % i == 0) return false;
  return true;
}
int dfs_flow(int u, int v, int f = I) {
  if (u == v) return f;
  if (mark[u]) return 0;
  mark[u] = true;
  for (int i = head[u]; i != -1; i = prv[i]) {
    if (!cap[i]) continue;
    int x = dfs_flow(to[i], v, min(f, cap[i]));
    if (x) {
      cap[i] -= x;
      cap[i ^ 1] += x;
      return x;
    }
  }
  return 0;
}
int max_flow(int u, int v) {
  int f = 0, x;
  while (x = dfs_flow(u, v)) {
    memset(mark, false, sizeof mark);
    f += x;
  }
  return f;
}
vector<int> table;
void dfs(int v) {
  mark[v] = true;
  table.push_back(v);
  for (int u : adj[v])
    if (!mark[u]) dfs(u);
}
bool valid(int x) { return x != N - 1 && x != N - 2; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  memset(head, -1, sizeof head);
  int n;
  cin >> n;
  vector<int> L, R;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i] % 2 == 0)
      L.push_back(i);
    else
      R.push_back(i);
  }
  if (L.size() != R.size()) shit();
  for (int i = 0; i < L.size(); ++i)
    for (int j = 0; j < R.size(); ++j)
      if (is_prime(a[L[i]] + a[R[j]])) add_edge(L[i], R[j], 1);
  for (int i = 0; i < L.size(); ++i) add_edge(N - 2, L[i], 2);
  for (int i = 0; i < R.size(); ++i) add_edge(R[i], N - 1, 2);
  int f = max_flow(N - 2, N - 1);
  if (f != n) shit();
  for (int x = 0; x < L.size(); ++x) {
    int u = L[x];
    for (int i = head[u]; i != -1; i = prv[i])
      if (!cap[i]) {
        int v = to[i];
        adj[u].push_back(v);
        adj[v].push_back(u);
      }
  }
  memset(mark, false, sizeof mark);
  int ans = 0;
  vector<int> out;
  for (int i = 0; i < L.size(); ++i) {
    if (mark[L[i]]) continue;
    table.clear();
    dfs(L[i]);
    ans++;
    out.push_back(table.size());
    for (int fox : table) out.push_back(fox + 1);
    out.push_back(-1);
  }
  cout << ans << "\n";
  for (int i = 0; i < out.size(); ++i)
    if (out[i] == -1)
      cout << "\n";
    else
      cout << out[i] << " ";
}
