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 n;
vector<int> A, B, aa, bb;
int q[555][555];
int in[555][555];
int second, T;
int f[555];
void er() {
  puts("Impossible");
  exit(0);
}
bool prime(int x) {
  for (int i = 2; i < x; i++)
    if (x % i == 0) return 0;
  return 1;
}
void ae(int x, int y, int f) {
  q[x][y] += f;
  in[x][y] += f;
}
void rec(int x) {
  for (int i = 0; i <= T; i++)
    if (q[x][i] > 0 && f[i] == -1) {
      f[i] = x;
      rec(i);
    }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    if (x % 2) {
      A.push_back(x);
      aa.push_back(i);
    } else {
      B.push_back(x);
      bb.push_back(i);
    }
  }
  if (A.size() != B.size()) er();
  second = 0;
  T = 1 + A.size() + B.size();
  for (int i = 0; i < A.size(); i++) ae(second, i + 1, 2);
  for (int i = 0; i < A.size(); i++)
    for (int j = 0; j < B.size(); j++)
      if (prime(A[i] + B[j])) ae(i + 1, 1 + A.size() + j, 1);
  for (int i = 0; i < B.size(); i++) ae(1 + A.size() + i, T, 2);
  int fl = 0;
  for (;;) {
    for (int i = 0; i <= T; i++) f[i] = -1;
    rec(second);
    if (f[T] == -1) break;
    fl++;
    int x = T;
    while (x != second) {
      q[f[x]][x]--;
      q[x][f[x]]++;
      x = f[x];
    }
  }
  if (fl != A.size() * 2) er();
  vector<vector<int> > ans;
  for (int i = 0; i < A.size(); i++) f[i] = 0;
  for (int i = 0; i < A.size(); i++)
    if (!f[i]) {
      vector<int> t;
      int x = i;
      int pr = -1;
      for (;;) {
        f[x] = 1;
        t.push_back(aa[x]);
        int g = -1;
        for (int j = 0; j < B.size(); j++)
          if (q[x + 1][1 + A.size() + j] < in[x + 1][1 + A.size() + j] &&
              j != pr)
            g = j;
        t.push_back(bb[g]);
        pr = x;
        for (int j = 0; j < A.size(); j++)
          if (q[1 + j][1 + A.size() + g] < in[1 + j][1 + A.size() + g] &&
              j != pr)
            x = j;
        pr = g;
        if (x == i) break;
      }
      ans.push_back(t);
    }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].size();
    for (int j = 0; j < ans[i].size(); j++) printf(" %d", ans[i][j] + 1);
    puts("");
  }
  return 0;
}