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 int64_t M1 = 998244353;
const int64_t M2 = 1000000007;
mt19937 rng((uint64_t)chrono::steady_clock::now().time_since_epoch().count());
int64_t match[410];
int64_t vis[410];
vector<int64_t> node[410];
int64_t p[40010];
int64_t n;
bool dfs(int64_t u) {
  if (u < 0 || vis[u]) return false;
  vis[u] = 1;
  for (int64_t v : node[u])
    if ((match[u + n] != v - n) && (match[v] == -1 || dfs(match[v]))) {
      match[v] = u;
      return true;
    }
  return false;
}
void solve() {
  p[0] = p[1] = 1;
  for (int64_t i = 2; i * i <= 40009; i++)
    if (!p[i])
      for (int64_t j = i + i; j <= 40009; j += i) p[j] = 1;
  cin >> n;
  vector<int64_t> a(n);
  for (int64_t i = 0; i < n; i++) cin >> a[i];
  for (int64_t i = 0; i < n; i++)
    for (int64_t j = 0; j < n; j++) {
      if (!p[a[i] + a[j]]) node[i].push_back(n + j);
    }
  memset(match, -1, sizeof(match));
  for (int64_t i = 0; i < n; i++) {
    memset(vis, 0, sizeof(vis));
    if (!dfs(i)) {
      cout << "Impossible\n";
      return;
    }
  }
  memset(vis, 0, sizeof(vis));
  vector<vector<int64_t>> tables;
  for (int64_t i = 0; i < n; i++) {
    if (!vis[i]) {
      vector<int64_t> cycle;
      cycle.push_back(i);
      vis[i] = 1;
      int64_t u = match[n + i];
      while (u != i) {
        vis[u] = 1;
        cycle.push_back(u);
        u = match[n + u];
      }
      tables.push_back(cycle);
    }
  }
  cout << (int64_t)tables.size() << "\n";
  for (auto cycle : tables) {
    cout << (int64_t)cycle.size() << " ";
    for (int64_t u : cycle) cout << u + 1 << " ";
    cout << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}