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;
const int N = 400 + 7;
const int M = 1e5 + 2;
const int INF = 1e9 + 7;
const int MAX = 1e5 + 7;
int head[N], prv[2 * M], from[2 * M], to[2 * M], cap[2 * M];
int cnte = 0;
bool mark[N], flag[N];
int n, m;
int a[N];
bool np[MAX];
int mat[N][N];
vector<int> adj[N];
vector<int> c[N];
int ans = 0;
void DFS(int v, vector<int> &c) {
  flag[v] = 1;
  c.push_back(v);
  for (auto u : adj[v])
    if (!flag[u]) DFS(u, c);
}
void add_edge(int v, int u, int vu, int uv = 0) {
  from[cnte] = v, to[cnte] = u, cap[cnte] = vu, prv[cnte] = head[v],
  head[v] = cnte++;
  from[cnte] = u, to[cnte] = v, cap[cnte] = uv, prv[cnte] = head[u],
  head[u] = cnte++;
}
int dfs(int v, int sink, int f = INF) {
  if (v == sink) return f;
  if (mark[v]) return 0;
  mark[v] = true;
  for (int i = head[v]; i != -1; i = prv[i])
    if (cap[i]) {
      int x = dfs(to[i], sink, min(cap[i], f));
      if (x) {
        cap[i] -= x;
        cap[i ^ 1] += x;
        return x;
      }
    }
  return 0;
}
int flow(int s, int t) {
  int out = 0, x;
  memset(mark, 0, sizeof mark);
  while (x = dfs(s, t)) {
    memset(mark, 0, sizeof mark);
    out += x;
  }
  return out;
}
int main() {
  ios::sync_with_stdio(false);
  memset(head, -1, sizeof head);
  np[0] = np[1] = 1;
  for (int i = 2; i < MAX; i++)
    if (!np[i])
      for (int j = 2 * i; j < MAX; j += i) np[j] = 1;
  int source = N - 1, sink = N - 2;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] % 2 == 0)
      add_edge(source, i, 2);
    else
      add_edge(i, sink, 2);
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (!np[a[i] + a[j]] && a[i] % 2 == 0) {
        mat[i][j] = cnte;
        add_edge(i, j, 1);
      }
  if (flow(source, sink) != n) {
    cout << "Impossible\n";
    return 0;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (!np[a[i] + a[j]] && a[i] % 2 == 0 && cap[mat[i][j]] == 0) {
        adj[i].push_back(j);
        adj[j].push_back(i);
      }
  for (int i = 0; i < n; i++)
    if (!flag[i]) {
      DFS(i, c[ans]);
      ans++;
    }
  cout << ans << "\n";
  for (int i = 0; i < ans; i++) {
    cout << c[i].size() << " ";
    for (auto x : c[i]) cout << x + 1 << " ";
    cout << "\n";
  }
  return 0;
}