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;
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& p) {
  return o << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vc) {
  o << "{";
  for (const T& v : vc) o << v << ",";
  o << "}";
  return o;
}
struct timer {
  double t;
  timer() { t = clock(); }
  ~timer() { printf("\nruntime %.3f s\n", get_time()); }
  double get_time() { return (clock() - t) / CLOCKS_PER_SEC; }
};
struct FlowEdge {
  long long v, u;
  long long cap, flow = 0;
  FlowEdge(long long v, long long u, long long cap) : v(v), u(u), cap(cap) {}
};
struct Dinic {
  const long long flow_inf = 1e18;
  vector<FlowEdge> edges;
  vector<vector<long long>> adj;
  long long n, m = 0;
  long long s, t;
  vector<long long> level, ptr;
  queue<long long> q;
  Dinic(long long n, long long s, long long t) : n(n), s(s), t(t) {
    adj.resize(n);
    level.resize(n);
    ptr.resize(n);
  }
  void add_edge(long long v, long long u, long long cap) {
    edges.push_back(FlowEdge(v, u, cap));
    edges.push_back(FlowEdge(u, v, 0));
    adj[v].push_back(m);
    adj[u].push_back(m + 1);
    m += 2;
  }
  bool bfs() {
    while (!q.empty()) {
      long long v = q.front();
      q.pop();
      for (long long id : adj[v]) {
        if (edges[id].cap - edges[id].flow < 1) continue;
        if (level[edges[id].u] != -1) continue;
        level[edges[id].u] = level[v] + 1;
        q.push(edges[id].u);
      }
    }
    return level[t] != -1;
  }
  long long dfs(long long v, long long pushed) {
    if (pushed == 0) return 0;
    if (v == t) return pushed;
    for (long long& cid = ptr[v]; cid < (long long)adj[v].size(); cid++) {
      long long id = adj[v][cid];
      long long u = edges[id].u;
      if (level[v] + 1 != level[u] || edges[id].cap - edges[id].flow < 1)
        continue;
      long long tr = dfs(u, min(pushed, edges[id].cap - edges[id].flow));
      if (tr == 0) continue;
      edges[id].flow += tr;
      edges[id ^ 1].flow -= tr;
      return tr;
    }
    return 0;
  }
  long long flow() {
    long long f = 0;
    while (true) {
      fill(level.begin(), level.end(), -1);
      level[s] = 0;
      q.push(s);
      if (!bfs()) break;
      fill(ptr.begin(), ptr.end(), 0);
      while (long long pushed = dfs(s, flow_inf)) {
        f += pushed;
      }
    }
    return f;
  }
};
long long n, a[500], e, vis[500];
std::vector<long long> adj[500], ans[500];
bool isp(long long x) {
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
long long cnt = 0;
void dfs1(long long s, long long p) {
  if (vis[s]) {
    return;
  }
  vis[s] = 1;
  ans[cnt].push_back(s);
  for (auto t : adj[s]) {
    if (t != p) {
      dfs1(t, s);
    }
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  Dinic d(n + 2, 0, n + 1);
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] % 2 == 0) {
      e++;
      d.add_edge(0, i, 2);
    } else {
      d.add_edge(i, n + 1, 2);
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      if (a[i] % 2 == 0) {
        if (isp(a[i] + a[j])) {
          d.add_edge(i, j, 1);
        }
      }
    }
  }
  long long f = d.flow();
  if (2 * e != f) {
    cout << "Impossible" << endl;
    return 0;
  }
  for (auto t : d.edges) {
    if (t.flow > 0 && t.u != 0 && t.v != 0 && t.u != n + 1 && t.v != n + 1) {
      adj[t.v].push_back(t.u);
      adj[t.u].push_back(t.v);
    }
  }
  cnt = 0;
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) {
      cnt++;
      dfs1(i, 0);
    }
  }
  cout << cnt << endl;
  for (long long i = 1; i <= cnt; i++) {
    cout << ans[i].size() << " ";
    for (auto t : ans[i]) {
      cout << t << " ";
    }
    cout << endl;
  }
  return 0;
}