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 inft = 1000000009;
const int MAXN = 210;
const int MAXM = 20100;
bool sieve[MAXM];
namespace TM {
const int MAXN = 210;
vector<int> V[MAXN];
int c, d;
int M[MAXN];
bool odw[MAXN];
queue<int> Q;
inline bool dfs(int u) {
  odw[u] = 1;
  Q.push(u);
  for (typeof((V[u]).begin()) it = (V[u]).begin(); it != (V[u]).end(); ++it)
    if ((M[*it] == -1) || (!odw[M[*it]] && dfs(M[*it]))) {
      M[u] = *it;
      M[*it] = u;
      return 1;
    }
  return 0;
}
int matching() {
  for (int i = 0; i < (c); ++i) odw[i] = 0;
  for (int i = 0; i < (c + d); ++i) M[i] = -1;
  bool czy = 1;
  while (czy) {
    for (czy = 0; !Q.empty(); Q.pop()) odw[Q.front()] = 0;
    for (int i = 0; i < (c); ++i)
      if (M[i] == -1 && !odw[i]) czy |= dfs(i);
  }
  int ret = 0;
  for (int i = 0; i < (c); ++i)
    if (M[i] != -1) ++ret;
  return ret;
}
}  // namespace TM
using TM::c;
using TM::d;
using TM::M;
using TM::V;
int MM[MAXN];
vector<int> vis;
void mdfs(vector<int> &cyc, int u) {
  if (vis[u]) return;
  cyc.push_back(u);
  vis[u] = 1;
  if (u < c)
    return mdfs(cyc, MM[u]);
  else
    return mdfs(cyc, M[u]);
}
void imp() {
  puts("Impossible");
  exit(0);
}
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> X(n), renum(n);
  for (int i = 0; i < (n); ++i) {
    int a;
    scanf("%d", &a);
    if (a % 2) {
      renum[i] = c++;
    } else {
      renum[i] = n - (++d);
    }
    X[renum[i]] = a;
  }
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (n); ++j)
      if (i < j) {
        if (!sieve[X[i] + X[j]]) {
          V[i].push_back(j);
          V[j].push_back(i);
          assert(i < c && j >= c);
        }
      }
  if (TM::matching() * 2 != c + d) imp();
  vector<vector<int> > W(n);
  for (int i = 0; i < (c); ++i)
    for (typeof((V[i]).begin()) it = (V[i]).begin(); it != (V[i]).end(); ++it) {
      if (M[i] != *it) {
        W[i].push_back(*it);
        W[*it].push_back(i);
      }
    }
  for (int i = 0; i < (n); ++i) swap(V[i], W[i]);
  for (int i = 0; i < (n); ++i) MM[i] = M[i];
  if (TM::matching() * 2 != c + d) imp();
  vector<vector<int> > ans;
  vis.resize(n);
  for (int i = 0; i < (n); ++i) {
    if (vis[i]) continue;
    vector<int> cyc;
    mdfs(cyc, i);
    assert(cyc.size() > 2);
    ans.push_back(cyc);
  }
  vector<int> odw(n);
  for (int i = 0; i < (n); ++i) odw[renum[i]] = i;
  printf("%lu\n", ans.size());
  for (typeof((ans).begin()) ic = (ans).begin(); ic != (ans).end(); ++ic) {
    printf("%lu ", ic->size());
    for (typeof((*ic).begin()) it = (*ic).begin(); it != (*ic).end(); ++it)
      printf("%d ", odw[*it] + 1);
    puts("");
  }
}
int main() {
  sieve[0] = sieve[1] = 1;
  for (int i = 2; i * i < MAXM; i++) {
    if (!sieve[i])
      for (int j = i * i; j < MAXM; j += i) sieve[j] = 1;
  }
  int t = 1;
  for (int i = 0; i < (t); ++i) solve();
  return 0;
}