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 cap[500][500];
int par[500];
bool vis[500];
int fl[500];
int f[500][500];
int a[500];
int odd, even;
int s, t, n;
vector<int> ans;
bool prime(int x) {
  if (x == 1) return false;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
queue<int> q;
bool bfs() {
  memset(par, -1, sizeof(par));
  par[s] = s;
  fl[s] = 1000007;
  q.push(s);
  while (q.size()) {
    int v = q.front();
    q.pop();
    for (int i = 1; i <= n + 1; i++) {
      if ((cap[v][i] > 0) && (par[i] == -1)) {
        par[i] = v;
        fl[i] = min(fl[v], cap[v][i]);
        q.push(i);
      }
    }
  }
  if (par[t] == -1) return false;
  int p = t;
  while (p != s) {
    cap[p][par[p]] += fl[t];
    cap[par[p]][p] -= fl[t];
    f[par[p]][p] += fl[t];
    f[p][par[p]] -= fl[t];
    p = par[p];
  }
  return true;
}
int max_flow() {
  int ans = 0;
  while (bfs())
    ;
  for (int i = 0; i < n + 2; i++) ans += f[i][t];
  return ans;
}
stringstream ss;
void trace(int v) {
  vis[v] = true;
  ans.push_back(v);
  for (int i = 1; i <= n; i++)
    if ((f[v][i] || f[i][v]) && !vis[i]) trace(i);
}
int main() {
  scanf("%d", &n);
  memset(f, 0, sizeof(f));
  memset(cap, 0, sizeof(cap));
  memset(fl, 0, sizeof(fl));
  s = 0;
  t = n + 1;
  int i, j;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] % 2) {
      cap[0][i] = 2;
      odd++;
    } else {
      cap[i][n + 1] = 2;
      even++;
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (a[i] % 2 == 1 && a[j] % 2 == 0 && prime(a[i] + a[j])) cap[i][j] = 1;
    }
  }
  int maxx = max_flow();
  if ((maxx != n) || (odd != even)) {
    printf("Impossible\n");
    return 0;
  }
  int count = 0;
  memset(vis, false, sizeof(vis));
  for (int i1 = 1; i1 <= n; i1++) {
    if (!vis[i1]) {
      count++;
      ans.clear();
      trace(i1);
      ss << ans.size() << " ";
      for (i = 0; i < ans.size(); i++) ss << ans[i] << " ";
      ss << "\n";
    }
  }
  cout << count << '\n';
  for (int i = 0; i < count; i++) {
    int x;
    ss >> x, cout << x << ' ';
    for (int i = 0, v; i < x; i++) ss >> v, cout << v << ' ';
    cout << '\n';
  }
  return 0;
}