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 mark[100000];
int cap[203][203], f[203][203];
int val[203], par[203];
int N, S = 201, T = 202;
int maxflow = 0;
int odd = 0, even = 0;
int vis[203];
int INF = 1234567;
vector<vector<int> > adj;
vector<vector<int> > temp;
int augment_path(int v, int mf) {
  int u = par[v];
  if (v == S) {
    return mf;
  }
  bool backedge = false;
  if (cap[u][v] - f[u][v] <= 0 && f[v][u] > 0) backedge = true;
  if (backedge) {
    mf = min(mf, f[v][u]);
  } else {
    mf = min(mf, cap[u][v] - f[u][v]);
  }
  mf = augment_path(u, mf);
  if (backedge) {
    f[v][u] -= mf;
  } else {
    f[u][v] += mf;
  }
  return mf;
}
void edmondskarp() {
  maxflow = 0;
  while (1) {
    bool augmented = false;
    for (int i = 0; i < 203; ++i) vis[i] = 0;
    queue<int> q;
    q.push(S);
    vis[S] = 1;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int i = 0; i < 203; ++i) {
        if ((cap[u][i] - f[u][i] > 0 || f[i][u] > 0) && !vis[i]) {
          vis[i] = 1;
          par[i] = u;
          if (i == T) {
            maxflow += augment_path(T, INF);
            augmented = true;
            break;
          } else
            q.push(i);
        }
      }
      if (augmented) break;
    }
    if (!augmented) break;
  }
}
void dfs(int u) {
  vis[u] = 1;
  temp.back().push_back(u);
  for (int i = 0; i < adj[u].size(); ++i) {
    if (vis[adj[u][i]]) continue;
    dfs(adj[u][i]);
  }
}
int main() {
  for (int i = 2; i * i < 100000; ++i) {
    for (int j = i; j * i < 100000; ++j) {
      mark[i * j] = 1;
    }
  }
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    scanf("%d", &val[i]);
  }
  for (int i = 0; i < N; ++i) {
    if (val[i] % 2) {
      cap[S][i] = 2;
      ++odd;
    } else {
      cap[i][T] = 2;
      ++even;
    }
    for (int j = i + 1; j < N; ++j) {
      if (mark[val[i] + val[j]]) continue;
      if (val[i] % 2)
        cap[i][j] = 1;
      else
        cap[j][i] = 1;
    }
  }
  edmondskarp();
  if (odd * 2 != even * 2 || even * 2 != maxflow) {
    printf("Impossible\n");
    return 0;
  }
  adj = vector<vector<int> >(N + 3);
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < N; ++j) {
      if (cap[i][j] > 0 && f[i][j] > 0) {
        adj[i].push_back(j);
        adj[j].push_back(i);
      }
    }
  }
  for (int i = 0; i < N; ++i) vis[i] = 0;
  for (int i = 0; i < N; ++i) {
    if (vis[i]) continue;
    temp.push_back(vector<int>());
    dfs(i);
  }
  printf("%d\n", (int)temp.size());
  for (int i = 0; i < temp.size(); ++i) {
    printf("%d", (int)temp[i].size());
    for (int j = 0; j < temp[i].size(); ++j) {
      printf(" %d", temp[i][j] + 1);
    }
    printf("\n");
  }
  return 0;
}