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 NMAX = 210, INF = 0x3f3f3f3f, VMAX = 20010;
int N, A[NMAX], Source, Sink, Flow[NMAX][NMAX], Cap[NMAX][NMAX], Father[NMAX],
    Even, Odd;
bool Used[NMAX], Prime[VMAX];
vector<int> G[NMAX], FlowG[NMAX], CurrentTable;
vector<vector<int> > Ans;
bool BFS() {
  queue<int> Q;
  for (int i = 0; i <= N + 1; ++i) Used[i] = Father[i] = 0;
  Used[Source] = 1;
  Q.push(Source);
  while (!Q.empty()) {
    int Node = Q.front();
    Q.pop();
    if (Node == Sink) return 1;
    for (int i = 0; i < G[Node].size(); ++i) {
      int Vec = G[Node][i];
      if (!Used[Vec] && Cap[Node][Vec] > Flow[Node][Vec])
        Used[Vec] = 1, Father[Vec] = Node, Q.push(Vec);
    }
  }
  return Used[Sink];
}
void DFS(int Node, int StartNode) {
  if (Node == StartNode && CurrentTable.size() >= 3) return;
  CurrentTable.push_back(Node);
  Used[Node] = 1;
  for (int i = 0; i < FlowG[Node].size(); ++i)
    if (!Used[FlowG[Node][i]]) DFS(FlowG[Node][i], StartNode);
}
void BuildGraph() {
  Source = 0;
  Sink = N + 1;
  for (int i = 1; i <= N; ++i)
    if (A[i] % 2 == 1) {
      G[Source].push_back(i);
      G[i].push_back(Source);
      Cap[Source][i] = 2;
    } else {
      G[i].push_back(Sink);
      G[Sink].push_back(i);
      Cap[i][Sink] = 2;
    }
  for (int i = 1; i <= N; ++i)
    for (int j = 1; j <= N; ++j)
      if (A[i] % 2 == 1 && A[j] % 2 == 0 && Prime[A[i] + A[j]]) {
        G[i].push_back(j);
        G[j].push_back(i);
        Cap[i][j] = 1;
      }
}
void MaxFlow() {
  int CntFlow = 0;
  while (BFS()) {
    for (int i = 0; i < G[Sink].size(); ++i) {
      int Vec = G[Sink][i];
      if (Used[Vec] && Cap[Vec][Sink] != Flow[Vec][Sink]) {
        int MinFlow = INF;
        Father[Sink] = Vec;
        for (int Node = Sink; Node != Source; Node = Father[Node])
          MinFlow =
              min(MinFlow, Cap[Father[Node]][Node] - Flow[Father[Node]][Node]);
        for (int Node = Sink; Node != Source; Node = Father[Node]) {
          Flow[Father[Node]][Node] += MinFlow;
          Flow[Node][Father[Node]] -= MinFlow;
        }
        CntFlow += MinFlow;
      }
    }
  }
  if (CntFlow != N)
    printf("Impossible\n");
  else {
    for (int i = 1; i <= N; ++i)
      for (int j = 1; j <= N; ++j)
        if (A[i] % 2 == 1 && A[j] % 2 == 0 && Prime[A[i] + A[j]] &&
            Flow[i][j] == 1)
          FlowG[i].push_back(j), FlowG[j].push_back(i);
    for (int i = 1; i <= N; ++i) Used[i] = 0;
    for (int i = 1; i <= N; ++i)
      if (!Used[i]) {
        CurrentTable.clear();
        DFS(i, i);
        Ans.push_back(CurrentTable);
      }
    printf("%i\n", Ans.size());
    for (int i = 0; i < Ans.size(); ++i) {
      printf("%i ", Ans[i].size());
      for (int j = 0; j < Ans[i].size(); ++j) printf("%i ", Ans[i][j]);
      printf("\n");
    }
    return;
  }
}
int main() {
  for (int i = 2; i < VMAX; ++i) Prime[i] = 1;
  for (int i = 2; i < VMAX; ++i)
    if (Prime[i])
      for (int j = i + i; j < VMAX; j += i) Prime[j] = 0;
  scanf("%i", &N);
  for (int i = 1; i <= N; ++i) {
    scanf("%i", &A[i]);
    Even += (A[i] % 2 == 0);
    Odd += (A[i] % 2 == 1);
  }
  if (Even != Odd) {
    printf("Impossible\n");
    return 0;
  }
  BuildGraph();
  MaxFlow();
  return 0;
}