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 ok[1 << 20];
int F[1 << 9][1 << 9];
int d[1 << 9];
int bfs(int n) {
  d[0] = 0;
  for (int i = 1; i <= n; ++i) d[i] = -1;
  queue<int> Q;
  Q.push(0);
  while (!Q.empty() && d[n] == -1) {
    int node = Q.front();
    Q.pop();
    for (int i = 0; i <= n; ++i)
      if (d[i] == -1 && F[node][i]) Q.push(i), d[i] = node;
  }
  if (d[n] == -1) return 0;
  int flow = 1e9;
  for (int vertex = n; vertex; vertex = d[vertex])
    flow = min(flow, F[d[vertex]][vertex]);
  for (int vertex = n; vertex; vertex = d[vertex])
    F[d[vertex]][vertex] -= flow, F[vertex][d[vertex]] += flow;
  return flow;
}
vector<int> order;
int mark[1 << 9];
void dfs(int node, int n, int how) {
  mark[node] = 1;
  order.push_back(node);
  for (int i = 1; i <= n; ++i)
    if (!mark[i] && (F[node][i] == how && F[i][node] != how))
      dfs(i, n, 1 - how);
}
int Val[1 << 9];
int main(void) {
  int n;
  for (int i = 2; i <= 1e6; ++i) ok[i] = 1;
  for (int i = 2; i <= 1e3; ++i)
    if (ok[i])
      for (int j = i * i; j <= 1e6; j += i) ok[j] = 0;
  cin >> n;
  vector<int> a, b, pa, pb;
  for (int i = 1; i <= n; ++i) {
    int val;
    cin >> val;
    Val[i] = val;
    if (val & 1)
      a.push_back(val), pa.push_back(i);
    else
      b.push_back(val), pb.push_back(i);
  }
  if (a.size() != b.size()) return puts("Impossible") * 0;
  int sz = a.size();
  for (int i = 1; i <= sz; ++i) F[0][i] = 2;
  for (int i = 1; i <= sz; ++i) F[i + sz][sz + sz + 1] = 2;
  for (int i = 0; i < sz; ++i)
    for (int j = 0; j < sz; ++j)
      if (ok[a[i] + b[j]]) F[i + 1][j + 1 + sz] = 1;
  int flow = 0;
  int cnt = 0;
  while ((cnt = bfs(sz + sz + 1)) != 0) flow += cnt;
  if (flow != sz + sz) return puts("Impossible") * 0;
  vector<vector<int> > cycles;
  for (int i = 1; i <= sz + sz; ++i)
    if (!mark[i]) {
      order.clear();
      dfs(i, sz + sz, 0);
      for (auto &it : order)
        if (it <= sz)
          it = pa[it - 1];
        else
          it = pb[it - sz - 1];
      cycles.push_back(order);
    }
  cout << (cycles.size()) << '\n';
  for (auto cycle : cycles) {
    cout << cycle.size() << ' ';
    for (auto it : cycle) cout << it << ' ';
    cout << '\n';
  }
  return 0;
}