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;
int res[300][300];
int sink;
int n;
int vis[300];
vector<int> graph[300];
int find_path(int u, int flow) {
  if (u == sink) return flow;
  vis[u] = 1;
  for (int i = 1; i <= sink; i++) {
    if (!vis[i] && res[u][i] > 0) {
      int f = find_path(i, min(flow, res[u][i]));
      if (f) {
        res[u][i] -= f;
        res[i][u] += f;
        return f;
      }
    }
  }
  return 0;
}
bool prime[200005];
void sieve() {
  for (int i = 2; i <= 2 * 10000 + 5; ++i) {
    if (prime[i] == 0) {
      for (int j = i * 2; j <= 2 * 10000 + 5; j += i) {
        prime[j] = 1;
      }
    }
  }
}
vector<int> ans[200];
void dfs(int c, int u) {
  ans[c].push_back(u);
  vis[u] = 1;
  for (int i = 0; i < graph[u].size(); i++) {
    int v = graph[u][i];
    if (!vis[v]) dfs(c, v);
  }
}
int v[300];
int main() {
  int n;
  sieve();
  scanf("%d", &n);
  int im = 0, pa = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
    if (v[i] & 1)
      im++;
    else
      pa++;
  }
  if (im != pa) {
    puts("Impossible");
    return 0;
  }
  sink = n + 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (prime[v[i] + v[j]] == 0 && (v[i] & 1)) {
        res[i][j] = 1;
      }
    }
    if (v[i] & 1) {
      res[0][i] = 2;
    } else {
      res[i][sink] = 2;
    }
  }
  int flow = 0;
  while (true) {
    memset(vis, 0, sizeof(vis));
    int a = find_path(0, (1 << 29));
    if (!a) break;
    flow += a;
  }
  if (flow != n) {
    puts("Impossible");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (v[i] & 1) continue;
    for (int j = 1; j <= n; j++) {
      if (res[i][j] == 1) {
        graph[i].push_back(j);
        graph[j].push_back(i);
      }
    }
  }
  memset(vis, 0, sizeof(vis));
  int k = 0;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    dfs(k++, i);
  }
  printf("%d\n", k);
  for (int i = 0; i < k; i++) {
    printf("%d ", (int)ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) {
      printf("%d ", ans[i][j]);
    }
    puts("");
  }
  return 0;
}
