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;
using namespace std;
const int INF = 2000000000;
struct Edge {
  int from, to, cap, flow, index;
  Edge(int from, int to, int cap, int flow, int index)
      : from(from), to(to), cap(cap), flow(flow), index(index) {}
};
struct Dinic {
  int N;
  vector<vector<Edge> > G;
  vector<Edge *> dad;
  vector<int> Q;
  Dinic(int N) : N(N), G(N), dad(N), Q(N) {}
  void AddEdge(int from, int to, int cap) {
    G[from].push_back(Edge(from, to, cap, 0, G[to].size()));
    if (from == to) G[from].back().index++;
    G[to].push_back(Edge(to, from, 0, 0, G[from].size() - 1));
  }
  long long BlockingFlow(int s, int t) {
    fill(dad.begin(), dad.end(), (Edge *)NULL);
    dad[s] = &G[0][0] - 1;
    int head = 0, tail = 0;
    Q[tail++] = s;
    while (head < tail) {
      int x = Q[head++];
      for (int i = 0; i < G[x].size(); i++) {
        Edge &e = G[x][i];
        if (!dad[e.to] && e.cap - e.flow > 0) {
          dad[e.to] = &G[x][i];
          Q[tail++] = e.to;
        }
      }
    }
    if (!dad[t]) return 0;
    long long totflow = 0;
    for (int i = 0; i < G[t].size(); i++) {
      Edge *start = &G[G[t][i].to][G[t][i].index];
      int amt = INF;
      for (Edge *e = start; amt && e != dad[s]; e = dad[e->from]) {
        if (!e) {
          amt = 0;
          break;
        }
        amt = min(amt, e->cap - e->flow);
      }
      if (amt == 0) continue;
      for (Edge *e = start; amt && e != dad[s]; e = dad[e->from]) {
        e->flow += amt;
        G[e->to][e->index].flow -= amt;
      }
      totflow += amt;
    }
    return totflow;
  }
  long long GetMaxFlow(int s, int t) {
    long long totflow = 0;
    while (long long flow = BlockingFlow(s, t)) totflow += flow;
    return totflow;
  }
};
bool isprime[100010];
void sieve(void) {
  fill(isprime, isprime + 100010, true);
  for (int i = 2; i < 100010; i++) {
    if (isprime[i]) {
      for (int j = 2 * i; j < 100010; j += i) isprime[j] = false;
    }
  }
}
int arr[100010];
int N;
vector<int> evens, odds;
vector<int> indices_even, indices_odd;
Dinic test(100010);
vector<int> neighbours[100010];
void FormGraph(void) {
  for (int i = 0; i < evens.size(); i++) {
    for (int j = 0; j < odds.size(); j++) {
      if (isprime[evens[i] + odds[j]])
        test.AddEdge(indices_even[i], indices_odd[j], 1);
    }
    test.AddEdge(0, indices_even[i], 2);
  }
  for (int i = 0; i < odds.size(); i++) test.AddEdge(indices_odd[i], N + 1, 2);
}
inline void ReadInput(void) {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] & 1)
      odds.push_back(arr[i]);
    else
      evens.push_back(arr[i]);
    if (arr[i] & 1)
      indices_odd.push_back(i);
    else
      indices_even.push_back(i);
  }
}
bool visited[100010];
vector<int> answers[100010];
inline void solve(void) {
  sieve();
  FormGraph();
  int max_matching = (int)test.GetMaxFlow(0, N + 1);
  int n1, n2;
  n1 = odds.size();
  n2 = evens.size();
  if ((max_matching != 2 * n1) or (max_matching != 2 * n2))
    cout << "Impossible";
  else {
    for (int i = 1; i <= N; i++) {
      for (int j = 0; j < test.G[i].size(); j++) {
        if (test.G[i][j].flow == 1) {
          neighbours[i].push_back(test.G[i][j].to);
        } else if (test.G[i][j].flow == -1) {
          neighbours[i].push_back(test.G[i][j].to);
        }
      }
    }
    fill(visited, visited + 100010, false);
    int counts = 0;
    for (int i = 1; i <= N; i++) {
      if (!visited[i]) {
        visited[i] = true;
        counts++;
        int curr = i;
        answers[counts].push_back(i);
        while (1) {
          int node1 = neighbours[curr][0];
          int node2 = neighbours[curr][1];
          if (visited[node1] and visited[node2]) break;
          int node = node1;
          if (visited[node]) node = node2;
          visited[node] = true;
          answers[counts].push_back(node);
          curr = node;
        }
      }
    }
    cout << counts << endl;
    for (int i = 1; i <= counts; i++) {
      cout << (int)answers[i].size() << " ";
      for (int j = 0; j < answers[i].size(); j++) cout << answers[i][j] << " ";
      cout << endl;
    }
  }
}
inline void Refresh(void) {}
int main() {
  ios_base::sync_with_stdio(false);
  ReadInput();
  solve();
  return 0;
}
