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.
#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)5e5 + 10;
const int mod = (int)1e9 + 7;
const int inf = (1 << 30) - 1;
int n;
int a[maxn];
int c[222][222];
int f[222][222];
int id[222];
int u[222];
int col[222];
int r[222];
int T;
int prime(int x) {
  for (int d = 2; d * d <= x; d++) {
    if (x % d == 0) return 0;
  }
  return 1;
}
bool dfs(int v) {
  if (v == n + 1) return 1;
  if (u[v] == T) return 0;
  u[v] = T;
  for (int i = 0; i <= n + 1; i++) {
    if (c[v][i] > f[v][i] && dfs(i)) {
      f[v][i]++;
      f[i][v]--;
      return 1;
    }
  }
  return 0;
}
vector<int> ord;
void dfs2(int v, int x) {
  col[v] = x;
  ord.emplace_back(r[v]);
  for (int i = 1; i <= n; i++) {
    if (col[i] == 0 && f[v][i] != 0) {
      dfs2(i, x);
    }
  }
}
void solve() {
  cin >> n;
  vector<int> e, o;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] % 2 == 0) {
      e.emplace_back(a[i]);
      id[i] = e.size();
      r[e.size()] = i;
    } else {
      o.emplace_back(a[i]);
      id[i] = o.size() + n / 2;
      r[o.size() + n / 2] = i;
    }
  }
  if (e.size() != o.size()) {
    cout << "Impossible\n";
    return;
  }
  for (int i = 0; i < (int)e.size(); i++) {
    for (int j = 0; j < (int)o.size(); j++) {
      if (prime(e[i] + o[j])) {
        c[i + 1][j + n / 2 + 1] = 1;
      }
    }
  }
  for (int i = 1; i <= n / 2; i++) {
    c[0][i] = 2;
    c[i + n / 2][n + 1] = 2;
  }
  int ans = 0;
  T++;
  while (dfs(0)) {
    T++;
    ans++;
  }
  if (ans != n) {
    cout << "Impossible\n";
    return;
  }
  int sz = 0;
  vector<vector<int> > A;
  for (int i = 1; i <= n; i++) {
    if (col[i] > 0) continue;
    ++sz;
    ord.clear();
    dfs2(i, sz);
    A.emplace_back(ord);
  }
  cout << sz << "\n";
  for (auto d : A) {
    cout << d.size();
    for (int v : d) {
      cout << " " << v + 1;
    }
    cout << "\n";
  }
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
