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 Prime[3 * 10100], capacity[210][210], N, a[210], parent[210],
    f_graph[210][210];
vector<int> adj[210];
bool vis[210];
void sieve() {
  for (int i = 0; i < 3 * 10100; i++) Prime[i] = true;
  Prime[0] = Prime[1] = false;
  for (int i = 2; i * i < 3 * 10100; i++)
    if (Prime[i])
      for (int j = i * i; j < 3 * 10100; j += i) Prime[j] = false;
}
int bfs(int s, int t) {
  for (int i = 1; i <= N; i++) parent[i] = -1;
  parent[s] = -2, parent[t] = -1;
  queue<pair<int, int> > Queue;
  Queue.push(make_pair(s, 100000000));
  while (!Queue.empty()) {
    int node = Queue.front().first, flow = Queue.front().second;
    Queue.pop();
    for (auto x : adj[node])
      if (parent[x] == -1 and capacity[node][x] > 0) {
        flow = min(flow, capacity[node][x]);
        parent[x] = node;
        if (x == t) return flow;
        Queue.push(make_pair(x, flow));
      }
  }
  return 0;
}
int maxflow(int s, int t) {
  int ret = 0;
  int tmp = bfs(s, t);
  while (tmp != 0) {
    ret += tmp;
    int p = t;
    while (p != s) {
      int prev = parent[p];
      if (p == t or prev == s or (a[p] & 1 and !(a[prev] & 1) and prev != t))
        f_graph[prev][p] += tmp;
      else
        f_graph[p][prev] -= tmp;
      capacity[prev][p] -= tmp;
      capacity[p][prev] += tmp;
      p = prev;
    }
    tmp = bfs(s, t);
  }
  return ret;
}
void dfs(int s, int p, int original, vector<int> &v) {
  if (p != -1 and s == original) return;
  vis[s] = true;
  v.push_back(s);
  for (auto x : adj[s])
    if (x != p and (f_graph[s][x] == 1 or f_graph[x][s] == 1)) {
      dfs(x, s, original, v);
      break;
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  sieve();
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> a[i];
  int cnt_odd = 0, cnt_even = 0;
  for (int i = 1; i <= N; i++)
    if (a[i] & 1)
      cnt_odd++;
    else
      cnt_even++;
  if (cnt_even != cnt_odd) {
    cout << "Impossible\n";
    return 0;
  }
  for (int i = 1; i <= N; i++)
    if (!(a[i] & 1))
      for (int j = 1; j <= N; j++)
        if (Prime[a[i] + a[j]])
          adj[i].push_back(j), adj[j].push_back(i), capacity[i][j] = 1;
  int s = 0, t = N + 1;
  for (int i = 1; i <= N; i++)
    if (a[i] & 1)
      adj[i].push_back(t), adj[t].push_back(i), capacity[i][t] = 2;
    else
      adj[s].push_back(i), adj[i].push_back(s), capacity[s][i] = 2;
  int flow = maxflow(s, t);
  if (flow != 2 * cnt_odd) {
    cout << "Impossible\n";
    return 0;
  }
  vector<vector<int> > cycle_list;
  for (int i = 1; i <= N; i++)
    if (!vis[i]) {
      vector<int> v;
      dfs(i, -1, i, v);
      cycle_list.push_back(v);
    }
  cout << cycle_list.size() << '\n';
  for (auto x : cycle_list) {
    cout << x.size() << " ";
    for (auto y : x) cout << y << " ";
    cout << '\n';
  }
  return 0;
}