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;
bool *primes_table(int last) {
  bool *table = new bool[last + 1];
  fill(table, table + last + 1, true);
  table[0] = false;
  table[1] = false;
  for (int i = 0; i < last + 1; i++)
    if (table[i])
      for (int j = i + i; j < last + 1; j += i) table[j] = false;
  return table;
}
int flow(int start, int end, int size, list<int> *graph, int **cap) {
  int answer = 0;
  int previous[size];
  bool used[size];
  while (true) {
    fill(used, used + size, false);
    fill(previous, previous + size, -1);
    queue<int> bfs;
    bfs.push(start);
    while (!bfs.empty()) {
      int cur = bfs.front();
      used[cur] = true;
      bfs.pop();
      for (int con : graph[cur])
        if (!used[con] and cap[cur][con] > 0) {
          previous[con] = cur;
          bfs.push(con);
        }
    }
    if (previous[end] == -1) break;
    int minimum = INT_MAX, cur = end;
    while (previous[cur] != -1) {
      minimum = min(cap[previous[cur]][cur], minimum);
      cur = previous[cur];
    }
    answer += minimum;
    cur = end;
    while (previous[cur] != -1) {
      cap[previous[cur]][cur] -= minimum;
      cap[cur][previous[cur]] += minimum;
      cur = previous[cur];
    }
  }
  return answer;
}
int n;
list<list<int>> table;
void dfs(int cur, bool *used, list<int> *graph, int **cap, int mode) {
  used[cur] = true;
  table.back().push_back(cur);
  for (int con : graph[cur])
    if (!used[con] and cap[cur][con] == mode)
      dfs(con, used, graph, cap, mode ^ 1);
}
int main() {
  scanf("%d", &n);
  int *cap[n + 2];
  list<int> graph[n + 2];
  bool *is_prime = primes_table(20000);
  vector<int> ages;
  for (int i = 0; i < n + 2; i++) {
    cap[i] = new int[n + 2];
    for (int j = 0; j < n + 2; j++) cap[i][j] = 0;
  }
  for (int i = 0; i < n; i++) {
    int age;
    scanf("%d", &age);
    if (age % 2 == 0) {
      graph[n].push_back(i);
      cap[n][i] = 2;
    } else {
      graph[i].push_back(n + 1);
      cap[i][n + 1] = 2;
    }
    for (int j = 0; j < ages.size(); j++)
      if (is_prime[ages[j] + age]) {
        graph[i].push_back(j);
        graph[j].push_back(i);
        if (age % 2 == 0)
          cap[i][j] = 1;
        else
          cap[j][i] = 1;
      }
    ages.push_back(age);
  }
  int result = flow(n, n + 1, n + 2, graph, cap);
  if (result != n) {
    printf("Impossible\n");
    return 0;
  }
  bool used[n + 2];
  used[n] = used[n + 1] = true;
  fill(used, used + n, false);
  for (int i = 0; i < n; i++) {
    if (used[i]) continue;
    table.push_back(list<int>());
    dfs(i, used, graph, cap, ages[i] % 2);
  }
  printf("%d\n", table.size());
  for (list<int> t : table) {
    printf("%d", t.size());
    for (int f : t) printf(" %d", f + 1);
    printf("\n");
  }
  return 0;
}