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;
void smain();
int main() {
  ios_base::sync_with_stdio(false);
  smain();
  return 0;
}
const long long N = 300;
long long a[N];
long long gr[N][N];
long long flw[N][N];
long long pred[N], fl[N], que[N];
long long n;
bool prime(long long a) {
  for (long long j = 2; j * j <= a; j++) {
    if (a % j == 0) return false;
  }
  return true;
}
vector<vector<long long> > ans;
void dfs(long long pos, long long q) {
  if (fl[pos] != 0) return;
  fl[pos] = 1;
  ans[q].push_back(pos);
  if (a[pos] % 2 == 0) {
    for (long long i = 0; i < n; i++) {
      if (gr[pos][i] > 0 && flw[pos][i] == 0 && fl[i] == 0) {
        dfs(i, q);
      }
    }
  } else {
    for (long long i = 0; i < n; i++) {
      if (gr[pos][i] > 0 && flw[i][pos] == 0 && fl[i] == 0) {
        dfs(i, q);
      }
    }
  }
}
void smain() {
  cin >> n;
  long long src = n;
  long long dst = n + 1;
  long long nn = n + 2;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n; i++) {
    if (a[i] % 2 == 0) {
      gr[src][i] = 2;
      flw[src][i] = 2;
      for (long long j = 0; j < n; j++) {
        if (a[j] % 2 == 1 && prime(a[i] + a[j])) {
          gr[i][j] = 1;
          gr[j][i] = 1;
          flw[i][j] = 1;
        }
      }
    } else {
      gr[i][dst] = 2;
      flw[i][dst] = 2;
    }
  }
  long long ll, rr, fr;
  while (true) {
    for (long long i = 0; i < nn; i++) {
      fl[i] = 0;
      pred[i] = -1;
    }
    ll = 0;
    rr = 0;
    que[0] = src;
    fl[src] = 1;
    while (ll <= rr) {
      fr = que[ll];
      for (long long i = 0; i < nn; i++) {
        if (flw[fr][i] > 0 && fl[i] == 0) {
          fl[i] = 1;
          rr++;
          que[rr] = i;
          pred[i] = fr;
        }
      }
      ll++;
    }
    if (fl[dst] == 0) break;
    ll = dst;
    rr = 10000;
    while (ll != src) {
      fr = pred[ll];
      rr = min(rr, flw[fr][ll]);
      ll = fr;
    }
    ll = dst;
    while (ll != src) {
      fr = pred[ll];
      flw[fr][ll] -= rr;
      flw[ll][fr] += rr;
      ll = fr;
    }
  }
  for (long long i = 0; i < nn; i++) {
    if (flw[src][i] != 0 || flw[i][dst] != 0) {
      cout << "Impossible";
      return;
    }
  }
  for (long long i = 0; i < n; i++) {
    fl[i] = 0;
  }
  long long q = 0;
  for (long long i = 0; i < n; i++) {
    if (fl[i] == 0 && a[i] % 2 == 0) {
      ans.push_back(vector<long long>());
      dfs(i, q);
      q++;
    }
  }
  cout << ans.size() << endl;
  for (long long i = 0; i < ans.size(); i++) {
    cout << ans[i].size();
    for (long long j = 0; j < ans[i].size(); j++) {
      cout << ' ' << (ans[i][j] + 1);
    }
    cout << '\n';
  }
}