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 inf = 1e9 + 333;
const long long linf = 1e18 + inf;
const int N = 200 + 5;
const int S = 200 + 1;
const int T = 200 + 2;
int n;
int a[N], d[N][N];
int prime(int x) {
  int sq = sqrt(x) + .5;
  for (int i = 2; i <= sq; i++)
    if (x % i == 0) return 0;
  return 1;
}
bool used[N];
bool dfs(int x) {
  if (x == T) return 1;
  used[x] = 1;
  for (int i = 1; i < N; i++) {
    if (!used[i] and d[x][i] and dfs(i)) {
      d[x][i]--;
      d[i][x]++;
      return 1;
    }
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    if (a[i] & 1) {
      d[S][i] = 2;
    } else {
      d[i][T] = 2;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1) {
      for (int j = 1; j <= n; j++) {
        if (prime(a[i] + a[j])) {
          d[i][j] = 1;
        }
      }
    }
  }
  int ans = 0;
  for (;;) {
    memset(used, 0, sizeof(used));
    if (!dfs(S)) break;
    ans++;
  }
  if (ans != n) {
    puts("Impossible");
    return 0;
  }
  memset(used, 0, sizeof(used));
  vector<vector<int> > ansV;
  for (int i = 1; i <= n; i++) {
    if (!used[i] and a[i] % 2) {
      vector<int> v;
      used[i] = 1;
      v.push_back(i);
      int x;
      for (int j = 1; j <= n; j++) {
        if (d[j][i]) {
          x = j;
          break;
        }
      }
      while (!used[x]) {
        used[x] = 1;
        v.push_back(x);
        for (int j = 1; j <= n; j++) {
          if (a[x] % 2) {
            if (!used[j] and d[j][x]) {
              x = j;
              break;
            }
          } else {
            if (!used[j] and d[x][j]) {
              x = j;
              break;
            }
          }
        }
      }
      ansV.push_back(v);
    }
  }
  printf("%d\n", ansV.size());
  for (auto v : ansV) {
    printf("%d", v.size());
    for (auto x : v) {
      printf(" %d", x);
    }
    puts("");
  }
  return 0;
}