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 MAXN = 202;
int c[MAXN][MAXN], f[MAXN][MAXN], in, a[MAXN], s, t;
int n;
bool mark[MAXN];
vector<int> ans[MAXN];
bool prime(int n) {
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
void make_graph() {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j && prime(a[i] + a[j]) && a[i] % 2 == 0) c[i][j] = 1;
  s = n, t = n + 1;
  for (int i = 0; i < n; i++)
    if (a[i] % 2 == 0) c[s][i] = 2;
  for (int i = 0; i < n; i++)
    if (a[i] % 2 == 1) c[i][t] = 2;
  n += 2;
}
int find_path(int n, int bn) {
  mark[n] = true;
  if (n == t) return bn;
  for (int i = 0; i < ::n; i++)
    if (!mark[i] && c[n][i] - f[n][i] > 0) {
      int val = find_path(i, min(bn, c[n][i] - f[n][i]));
      if (val > 0) {
        f[n][i] += val;
        f[i][n] -= val;
        return val;
      }
    }
  return 0;
}
int flow() {
  int F = 0;
  while (1) {
    fill(mark, mark + MAXN, 0);
    int val = find_path(s, 1e9);
    if (val == 0) return F;
    F += val;
  }
  return F;
}
void DFS(int n) {
  mark[n] = true;
  for (int i = 0; i < ::n; i++)
    if (!mark[i] && f[n][i]) DFS(i);
  ans[in].push_back(n);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  make_graph();
  int f = flow();
  n -= 2;
  if (f < n) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (!mark[i]) {
      DFS(i);
      in++;
    }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (::f[i][j]) ::f[j][i] = 1;
  cout << in << endl;
  for (int i = 0; i < in; i++, cout << endl) {
    cout << ans[i].size() << ' ';
    for (int j = 0; j < ans[i].size(); j++) cout << ans[i][j] + 1 << ' ';
  }
  return 0;
}