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;
class e {
 public:
  int v, c, f;
  e* rev;
  e(int v, int c) : v(v), c(c), f(0) {}
  bool pos() { return c > f; }
  void flow() {
    f++;
    rev->f--;
  }
};
vector<e*> con[210];
int n, a[201];
void addedge(int u, int v, int c) {
  e* edge = new e(v, c);
  e* rev = new e(u, 0);
  edge->rev = rev;
  rev->rev = edge;
  con[u].push_back(edge);
  con[v].push_back(rev);
}
int maxflow(int src, int snc) {
  int res = 0, par[210];
  e* trace[210];
  while (1) {
    memset(par, -1, sizeof(par));
    queue<int> q;
    q.push(src);
    while (!q.empty() && par[snc] < 0) {
      int u = q.front();
      q.pop();
      for (int i = 0; i < con[u].size(); i++) {
        int v = con[u][i]->v;
        if (par[v] < 0 && con[u][i]->pos()) {
          par[v] = u;
          trace[v] = con[u][i];
          q.push(v);
        }
      }
    }
    if (par[snc] < 0) return res;
    for (int cur = snc; cur != src; cur = par[cur]) trace[cur]->flow();
    res++;
  }
}
bool np[20001];
vector<int> c[210];
vector<int> res[210];
int gc;
bool visit[210];
void dfs(int u) {
  visit[u] = true;
  res[gc].push_back(u);
  for (int i = 0; i < c[u].size(); i++) {
    int v = c[u][i];
    if (!visit[v]) dfs(v);
  }
}
int main() {
  for (int i = 2; i <= 20000; i++) {
    if (!np[i]) {
      for (int j = 2; i * j <= 20000; j++) np[i * j] = true;
    }
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int src = n, snc = n + 1;
  for (int i = 0; i < n; i++) {
    if (a[i] % 2)
      addedge(src, i, 2);
    else
      addedge(i, snc, 2);
  }
  for (int i = 0; i < n; i++)
    if (a[i] % 2) {
      for (int j = 0; j < n; j++)
        if (a[j] % 2 == 0) {
          if (!np[a[i] + a[j]]) addedge(i, j, 1);
        }
    }
  if (maxflow(src, snc) != n) {
    puts("Impossible");
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (a[i] % 2) {
      for (int j = 0; j < con[i].size(); j++) {
        if (con[i][j]->f == 1) {
          c[i].push_back(con[i][j]->v);
          c[con[i][j]->v].push_back(i);
        }
      }
    }
  for (int i = 0; i < n; i++)
    if (!visit[i]) {
      dfs(i);
      gc++;
    }
  printf("%d\n", gc);
  for (int i = 0; i < gc; i++) {
    printf("%d ", (int)res[i].size());
    for (int j = 0; j < res[i].size(); j++) printf("%d ", res[i][j] + 1);
    puts("");
  }
  return 0;
}