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;
template <int SZ>
struct Dinic {
  struct Edge {
    int to, rev;
    long long flow, cap;
  };
  int N, s, t;
  vector<Edge> adj[SZ];
  typename vector<Edge>::iterator cur[SZ];
  void addEdge(int u, int v, long long cap) {
    assert(cap >= 0);
    Edge a{v, (int)adj[v].size(), 0, cap}, b{u, (int)adj[u].size(), 0, 0};
    adj[u].push_back(a), adj[v].push_back(b);
  }
  int level[SZ];
  bool bfs() {
    for (int i = 0; i < N; i++) level[i] = -1, cur[i] = begin(adj[i]);
    queue<int> q({s});
    level[s] = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (auto &e : adj[u])
        if (level[e.to] < 0 && e.flow < e.cap)
          q.push(e.to), level[e.to] = level[u] + 1;
    }
    return level[t] >= 0;
  }
  long long sendFlow(int v, long long flow) {
    if (v == t) return flow;
    for (; cur[v] != end(adj[v]); cur[v]++) {
      Edge &e = *cur[v];
      if (level[e.to] != level[v] + 1 || e.flow == e.cap) continue;
      auto df = sendFlow(e.to, min(flow, e.cap - e.flow));
      if (df) {
        e.flow += df;
        adj[e.to][e.rev].flow -= df;
        return df;
      }
    }
    return 0;
  }
  long long maxFlow(int n, int _s, int _t) {
    N = n, s = _s, t = _t;
    if (s == t) return -1;
    long long tot = 0;
    while (bfs())
      while (auto df = sendFlow(s, numeric_limits<long long>::max())) tot += df;
    return tot;
  }
};
int n, odd, use[502], resp;
vector<int> v, ans[502];
void dfs(Dinic<500> &g, int no, int comp, int pai = -1) {
  use[no] = 1;
  ans[comp].push_back(no);
  for (auto &ed : g.adj[no]) {
    if (ed.to < n && ed.to != pai && !use[ed.to]) {
      if (v[no] % 2 == 1 && ed.flow == 1) dfs(g, ed.to, comp, no);
      if (v[no] % 2 == 0 && ed.flow == -1) dfs(g, ed.to, comp, no);
    }
  }
}
bool is_prime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
int main() {
  scanf("%d", &n);
  Dinic<500> g;
  v.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
    odd += (v[i] % 2);
    if (v[i] % 2)
      g.addEdge(404, i, 2);
    else
      g.addEdge(i, 405, 2);
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (v[i] % 2 == 1 && v[j] % 2 == 0 && is_prime(v[i] + v[j]))
        g.addEdge(i, j, 1);
  int flow = g.maxFlow(500, 404, 405);
  if (2 * odd != n || flow != n) {
    puts("Impossible");
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (use[i]) continue;
    dfs(g, i, resp++);
  }
  printf("%d\n", resp);
  for (int i = 0; i < resp; i++) {
    printf("%d", ans[i].size());
    for (int x : ans[i]) printf(" %d", x + 1);
    puts("");
  }
  return 0;
}
