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 a[202];
bool vis[20000 + 10];
bool v[202];
vector<int> q[202];
int linkx[202], linky[202];
vector<int> ans[202];
int cnt;
bool dfs(int x) {
  for (int i = 0; i < q[x].size(); i++) {
    int to = q[x][i];
    if (v[to] || to == linky[x]) continue;
    v[to] = 1;
    if (!linky[to] || dfs(linky[to])) {
      linky[to] = x, linkx[x] = to;
      return 1;
    }
  }
  return 0;
}
int main() {
  for (int i = 2; i <= 20000; i++) {
    if (!vis[i]) {
      for (int j = i + i; j <= 20000; j += i) vis[j] = 1;
    }
  }
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i] + a[j] <= 20000 && !vis[a[i] + a[j]]) {
        q[i].push_back(j);
      }
    }
  }
  int ret = 0;
  for (int i = 1; i <= n; i++) {
    memset(v, 0, sizeof v);
    if (dfs(i)) ret++;
  }
  if (ret != n)
    printf("Impossible\n");
  else {
    memset(v, 0, sizeof v);
    for (int i = 1; i <= n; i++) {
      if (v[i]) continue;
      cnt++;
      int x = i;
      while (1) {
        v[x] = 1;
        ans[cnt].push_back(x);
        x = linkx[x];
        if (v[x]) break;
      }
    }
    printf("%d\n", cnt);
    for (int i = 1; i <= cnt; i++) {
      printf("%d ", ans[i].size());
      for (int j = 0; j < ans[i].size(); j++) {
        int to = ans[i][j];
        printf("%d ", to);
      }
      printf("\n");
    }
  }
  return 0;
}