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>
const double PI = 2.0 * acos(0.0);
using namespace std;
int n, a[250], even, odd, s, t, mf, f, res[250][250], level[250], p[250],
    set_size[250], disjoint_sets;
bool prime[20010], vis[250];
vector<int> adj[250];
void init() {
  memset(prime, -1, sizeof prime);
  for (int i = 3; i < 150; i++)
    if (prime[i])
      for (int j = i * i; j < 20010; j += i + i) prime[j] = false;
}
bool bfs() {
  queue<int> q;
  q.push(s);
  memset(level, -1, sizeof level);
  level[s] = 0;
  int u, v;
  while (q.size()) {
    u = q.front();
    q.pop();
    for (int i = 0; i < (int)adj[u].size(); i++) {
      v = adj[u][i];
      if (res[u][v] > 0 && level[v] == -1) {
        level[v] = level[u] + 1;
        q.push(v);
        if (v == t) return true;
      }
    }
  }
  return false;
}
int dfs(int u, int min_edge) {
  if (u == t) return min_edge;
  int v, faux, fsum = 0;
  for (int i = 0; i < (int)adj[u].size(); i++) {
    v = adj[u][i];
    if (res[u][v] > 0 && level[v] == level[u] + 1) {
      faux = dfs(v, min(min_edge, res[u][v]));
      if (faux) {
        fsum += faux;
        res[u][v] -= faux;
        res[v][u] += faux;
        min_edge -= faux;
        if (min_edge == 0) break;
      }
    }
  }
  return fsum;
}
int max_flow() {
  mf = 0;
  while (bfs()) mf += dfs(s, 2147483647);
  return mf;
}
int find_set(int i) { return (p[i] == i) ? i : (p[i] = find_set(p[i])); }
void union_set(int i, int j) {
  int x = find_set(i), y = find_set(j);
  if (x != y) {
    p[y] = x;
    set_size[x] += set_size[y];
    disjoint_sets--;
  }
}
void print(int u) {
  vis[u] = true;
  cout << " " << u;
  for (int i = 0; i < (int)adj[u].size(); i++) {
    int v = adj[u][i];
    if (!vis[v] && res[u][v] == ((a[u] % 2 == 0) ? 0 : 1)) print(v);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  int TC, i, j;
  init();
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] % 2 == 0)
      even++;
    else
      odd++;
  }
  if (even != odd)
    cout << "Impossible" << endl;
  else {
    s = 0;
    t = n + 1;
    for (i = 1; i <= n; i++) {
      if (a[i] % 2 != 0) {
        adj[i].push_back(t);
        adj[t].push_back(i);
        res[i][t] = 2;
      } else {
        adj[s].push_back(i);
        adj[i].push_back(s);
        res[s][i] = 2;
        for (j = 1; j <= n; j++) {
          if ((i != j) && (a[j] % 2 != 0) && prime[a[i] + a[j]]) {
            adj[i].push_back(j);
            adj[j].push_back(i);
            res[i][j] = 1;
          }
        }
      }
    }
    max_flow();
    if (mf != even * 2)
      cout << "Impossible" << endl;
    else {
      disjoint_sets = n;
      for (i = 1; i <= n; i++) p[i] = i, set_size[i] = 1;
      for (i = 1; i <= n; i++)
        if (a[i] % 2 == 0)
          for (j = 0; j < (int)adj[i].size(); j++)
            if ((adj[i][j] != s) && (res[i][adj[i][j]] == 0))
              union_set(i, adj[i][j]);
      cout << disjoint_sets << endl;
      memset(vis, 0, sizeof vis);
      vis[s] = vis[t] = true;
      for (i = 1; i <= n; i++)
        if (!vis[i]) cout << set_size[find_set(i)], print(i), cout << endl;
    }
  }
  return 0;
}