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 INF = 2000000000;
struct Edge {
  int from, to, cap, flow, index;
  Edge(int from, int to, int cap, int flow, int index)
      : from(from), to(to), cap(cap), flow(flow), index(index) {}
};
struct Dinic {
  int N;
  vector<vector<Edge> > G;
  vector<Edge *> dad;
  vector<int> Q;
  Dinic(int N) : N(N), G(N), dad(N), Q(N) {}
  void AddEdge(int from, int to, int cap) {
    G[from].push_back(Edge(from, to, cap, 0, G[to].size()));
    if (from == to) G[from].back().index++;
    G[to].push_back(Edge(to, from, 0, 0, G[from].size() - 1));
  }
  long long BlockingFlow(int s, int t) {
    fill(dad.begin(), dad.end(), (Edge *)NULL);
    dad[s] = &G[0][0] - 1;
    int head = 0, tail = 0;
    Q[tail++] = s;
    while (head < tail) {
      int x = Q[head++];
      for (int i = 0; i < G[x].size(); i++) {
        Edge &e = G[x][i];
        if (!dad[e.to] && e.cap - e.flow > 0) {
          dad[e.to] = &G[x][i];
          Q[tail++] = e.to;
        }
      }
    }
    if (!dad[t]) return 0;
    long long totflow = 0;
    for (int i = 0; i < G[t].size(); i++) {
      Edge *start = &G[G[t][i].to][G[t][i].index];
      int amt = INF;
      for (Edge *e = start; amt && e != dad[s]; e = dad[e->from]) {
        if (!e) {
          amt = 0;
          break;
        }
        amt = min(amt, e->cap - e->flow);
      }
      if (amt == 0) continue;
      for (Edge *e = start; amt && e != dad[s]; e = dad[e->from]) {
        e->flow += amt;
        G[e->to][e->index].flow -= amt;
      }
      totflow += amt;
    }
    return totflow;
  }
  long long GetMaxFlow(int s, int t) {
    long long totflow = 0;
    while (long long flow = BlockingFlow(s, t)) totflow += flow;
    return totflow;
  }
};
bool prime[20010];
void sieve() {
  int k = sqrt(20010);
  for (int i = 3; i <= k; i++) {
    if (!prime[i]) {
      for (int j = i * i; j < 20001; j += i + i) {
        prime[j] = 1;
      }
    }
  }
  return;
}
int a[210];
vector<int> adj[210];
vector<vector<int> > ans;
vector<int> tmp;
bool vis[210];
void dfs(int u) {
  vis[u] = 1;
  tmp.push_back(u);
  for (int i = 0; i < (int)adj[u].size(); i++) {
    int v = adj[u][i];
    if (!vis[v]) {
      dfs(v);
      return;
    }
  }
  return;
}
inline bool chk(int x, int n) { return (x >= 1 && x <= n); }
int main() {
  sieve();
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  Dinic din(n + 2);
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1) {
      din.AddEdge(0, i, 2);
    } else {
      din.AddEdge(i, n + 1, 2);
    }
    for (int j = 1; j <= n; j++) {
      if (a[i] % 2 == 1 && a[j] % 2 == 0 && !prime[a[i] + a[j]]) {
        din.AddEdge(i, j, 1);
      }
    }
  }
  if (din.GetMaxFlow(0, n + 1) != n) {
    cout << "Impossible\n";
  } else {
    for (int i = 0; i < (int)din.G.size(); i++) {
      for (int j = 0; j < (int)din.G[i].size(); j++) {
        if (din.G[i][j].flow && chk(din.G[i][j].from, n) &&
            chk(din.G[i][j].to, n)) {
          adj[din.G[i][j].from].push_back(din.G[i][j].to);
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) {
        (tmp.clear());
        dfs(i);
        ans.push_back(tmp);
      }
    }
    cout << (int)ans.size() << '\n';
    for (int i = 0; i < (int)ans.size(); i++) {
      cout << (int)ans[i].size();
      for (int j = 0; j < (int)ans[i].size(); j++) {
        cout << ' ' << ans[i][j];
      }
      cout << '\n';
    }
  }
  return 0;
}