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 = 200 + 10;
const int Maxa = 10000 + 10;
int n, ans;
int a[Maxn];
int adj[Maxn][Maxn];
bool isC[2 * Maxa];
bool mark[Maxn];
vector<vector<int> > cycle;
vector<int> c;
void input();
void find_p();
void init();
void flow();
bool dfs(int v);
void find_c();
void fc(int v);
void output();
int main() {
  input();
  find_p();
  init();
  flow();
  output();
}
void input() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  n++;
}
void find_p() {
  for (int i = 2; i <= 2 * 10000; i++)
    if (!isC[i])
      for (int j = 2 * i; j <= 2 * 10000; j += i) isC[j] = true;
}
void init() {
  for (int i = 1; i < n; i++)
    if (a[i] % 2 == 0)
      adj[0][i] = 2;
    else
      adj[i][n] = 2;
  for (int i = 1; i < n; i++)
    for (int j = 1; j < n; j++)
      if (a[i] % 2 == 0 && a[j] % 2 == 1 && !isC[a[i] + a[j]]) adj[i][j] = 1;
}
void flow() {
  while (dfs(0)) {
    ans++;
    for (int i = 0; i <= n; i++) mark[i] = false;
  }
}
bool dfs(int v) {
  mark[v] = true;
  if (v == n) return true;
  for (int i = 0; i <= n; i++)
    if (!mark[i] && adj[v][i] && dfs(i)) {
      adj[v][i]--;
      adj[i][v]++;
      return true;
    }
  return false;
}
void find_c() {
  for (int i = 0; i <= n; i++) mark[i] = false;
  for (int i = 1; i < n; i++)
    if (!mark[i]) {
      fc(i);
      cycle.push_back(c);
      c.clear();
    }
}
void fc(int v) {
  mark[v] = true;
  c.push_back(v);
  for (int i = 1; i < n; i++)
    if (((a[v] % 2 == 0 && adj[i][v]) || (a[v] % 2 == 1 && adj[v][i])) &&
        !mark[i])
      fc(i);
}
void output() {
  if (ans != n - 1)
    cout << "Impossible" << endl;
  else {
    find_c();
    cout << cycle.size() << endl;
    for (int i = 0; i < cycle.size(); i++) {
      cout << cycle[i].size() << " ";
      for (int j = 0; j < cycle[i].size(); j++) cout << cycle[i][j] << " ";
      cout << endl;
    }
  }
}