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 n, a[202], res[205][205], f, t, s;
bitset<20015> isprime;
bitset<205> vis;
vector<vector<int> > ans;
vector<int> p, q;
void augment(int v, int minEdge) {
  if (v == p[v]) return;
  if (v == s) {
    f = minEdge;
    return;
  }
  if (p[v] != -1) {
    augment(p[v], min(minEdge, res[p[v]][v]));
    if (f) q[p[v]] = v;
    res[v][p[v]] += f;
    res[p[v]][v] -= f;
  }
}
int main() {
  isprime.set();
  isprime[0] = isprime[1] = 0;
  for (int i = 0; i < 20015; i++)
    if (isprime[i]) {
      for (int j = i * i; j < 20015; j += i) isprime[j] = 0;
    }
  scanf("%d", &n);
  p.assign(n + 2, 0);
  q.assign(n + 2, 0);
  int cnt[] = {0, 0};
  s = n;
  t = n + 1;
  memset(res, 0, sizeof res);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    if (a[i] & 1) {
      cnt[0]++;
      res[s][i] = 2;
    } else {
      res[i][t] = 2;
      cnt[1]++;
    }
  }
  if (cnt[0] != cnt[1]) {
    puts("Impossible");
    return 0;
  }
  for (int i = 1; i < n; i++)
    for (int j = 0; j < i; j++)
      if (isprime[a[i] + a[j]]) {
        if (a[i] & 1)
          res[i][j] = 1;
        else
          res[j][i] = 1;
      }
  int flow = 0;
  n += 2;
  while (1) {
    f = 0;
    vis.reset();
    queue<int> q;
    p.assign(n, -1);
    q.push(s);
    vis[s] = 1;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      if (u == t) break;
      for (int i = 0; i < n; i++)
        if (!vis[i] && res[u][i] > 0) p[i] = u, vis[i] = 1, q.push(i);
    }
    augment(t, 1000000);
    if (f == 0) break;
    flow += f;
  }
  if (flow != (cnt[0] << 1)) {
    puts("Impossible");
    return 0;
  }
  vis.set();
  n -= 2;
  for (int i = 0; i < n; i++)
    if (vis[i] && (a[i] & 1)) {
      int v = i;
      ans.push_back(vector<int>());
      while (vis[v]) {
        ans.back().push_back(v);
        vis[v] = 0;
        for (int j = 0; j < n; j++) {
          if (vis[j] && ((a[j] & 1) == 0) && res[j][v]) {
            v = j;
            break;
          }
        }
        ans.back().push_back(v);
        vis[v] = 0;
        for (int j = 0; j < n; j++)
          if (vis[j] && (a[j] & 1) && res[v][j]) {
            v = j;
            break;
          }
      }
    }
  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] + 1);
    printf("\n");
  }
  return 0;
}
