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 = 203;
const int MAXN = N * 4;
int n;
struct MAXFLOW {
  const int INF = 100000000;
  struct edge {
    int a, b, cap, flow;
  };
  int s, t, d[MAXN], ptr[MAXN], q[MAXN];
  vector<edge> e;
  vector<int> g[MAXN];
  void add_edge(int a, int b, int cap) {
    edge e1 = {a, b, cap, 0};
    edge e2 = {b, a, 0, 0};
    g[a].push_back((int)e.size());
    e.push_back(e1);
    g[b].push_back((int)e.size());
    e.push_back(e2);
  }
  bool bfs() {
    int qh = 0, qt = 0;
    q[qt++] = s;
    memset(d, -1, sizeof(d));
    d[s] = 0;
    while (qh < qt && d[t] == -1) {
      int v = q[qh++];
      for (size_t i = 0; i < g[v].size(); ++i) {
        int id = g[v][i], to = e[id].b;
        if (d[to] == -1 && e[id].flow < e[id].cap) {
          q[qt++] = to;
          d[to] = d[v] + 1;
        }
      }
    }
    return d[t] != -1;
  }
  int dfs(int v, int flow) {
    if (!flow) return 0;
    if (v == t) return flow;
    for (; ptr[v] < (int)g[v].size(); ++ptr[v]) {
      int id = g[v][ptr[v]], to = e[id].b;
      if (d[to] != d[v] + 1) continue;
      int pushed = dfs(to, min(flow, e[id].cap - e[id].flow));
      if (pushed) {
        e[id].flow += pushed;
        e[id ^ 1].flow -= pushed;
        return pushed;
      }
    }
    return 0;
  }
  int dinic() {
    int flow = 0;
    for (;;) {
      if (!bfs()) break;
      memset(ptr, 0, sizeof(ptr));
      while (int pushed = dfs(s, INF)) flow += pushed;
    }
    return flow;
  }
} flow;
bool vis[N];
vector<int> adj[N];
vector<int> res;
void dfs(int v) {
  vis[v] = 1;
  res.push_back(v);
  for (auto u : adj[v])
    if (!vis[u]) dfs(u);
}
int a[N];
bool pr(int n) {
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  flow.s = n;
  flow.t = n + 1;
  int odd = 0, even = 0;
  for (int i = 0; i < n; i++)
    if (a[i] & 1)
      flow.add_edge(flow.s, i, 2), odd++;
    else
      flow.add_edge(i, flow.t, 2), even++;
  for (int i = 0; i < n; i++)
    if (a[i] & 1)
      for (int j = 0; j < n; j++)
        if (pr(a[i] + a[j])) {
          flow.add_edge(i, j, 1);
        }
  ::n = n + 3;
  int res = flow.dinic();
  if (res != odd * 2 || even != odd) return puts("Impossible"), 0;
  for (MAXFLOW::edge ed : flow.e) {
    if (ed.flow == 1) adj[ed.a].push_back(ed.b), adj[ed.b].push_back(ed.a);
  }
  vector<vector<int> > ans;
  for (int i = 0; i < n; i++)
    if (!vis[i]) {
      ::res.clear();
      dfs(i);
      ans.push_back(::res);
    }
  printf("%d\n", ans.size());
  for (auto v : ans) {
    printf("%d", v.size());
    for (auto x : v) printf(" %d", x + 1);
    puts("");
  }
  return 0;
}