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 a[410];
int g[410][410];
int vis[410], path[410];
queue<int> q;
int prime(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
int bfs(int s, int t) {
  for (int i = 0; i <= t; i++) {
    vis[i] = 0;
  }
  while (!q.empty()) q.pop();
  q.push(s), vis[s] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i <= t; i++) {
      if (vis[i] || g[u][i] <= 0) continue;
      vis[i] = 1, path[i] = u;
      if (i == t) break;
      q.push(i);
    }
  }
  if (!vis[t]) return 0;
  int now = t, minj = 0x3f3f3f3f;
  while (now != s) {
    int last = path[now];
    minj = min(minj, g[last][now]);
    now = last;
  }
  now = t;
  while (now != s) {
    int last = path[now];
    g[last][now] -= minj;
    g[now][last] += minj;
    now = last;
  }
  return minj;
}
int main() {
  int n;
  scanf("%d", &n);
  int s = 0, t = n + n + 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int tot = 0;
  for (int i = 1; i <= n; i++) {
    tot += 1 - a[i] % 2;
    for (int j = 1; j <= n; j++) {
      if (a[i] % 2 == 1 && a[j] % 2 != 1 && prime(a[i] + a[j])) g[i][j + n] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2)
      g[s][i] = 2;
    else
      g[i + n][t] = 2;
  }
  int flow = 0;
  while (1) {
    int tmp = bfs(s, t);
    if (!tmp) break;
    flow += tmp;
  }
  if (flow != tot * 2) {
    printf("Impossible\n");
    return 0;
  }
  vector<vector<int> > ans;
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    ans.push_back({});
    int now = i, last = now, pd = 1;
    while (pd || now != i) {
      pd = 0;
      ans.back().push_back(now);
      vis[now] = 1;
      for (int j = 1; j <= n; j++) {
        if (a[now] % 2 == a[j] % 2 || j == last) continue;
        if (a[now] % 2 == 1 && g[j + n][now] ||
            a[j] % 2 == 1 && g[now + n][j]) {
          last = now, now = j;
          break;
        }
      }
    }
  }
  if (ans.size() > n / 3) {
    printf("Impossible\n");
    return 0;
  }
  for (int i = 0; i < ans.size(); i++) {
    if (ans[i].size() < 3) {
      printf("Impossible\n");
      return 0;
    }
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) {
    printf("%d ", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) {
      printf("%d ", ans[i][j]);
    }
    printf("\n");
  }
  return 0;
}