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;
inline void CANNOT() { exit(0 * printf("Impossible\n")); }
const int MX = 2e4;
int n;
int a[205];
vector<int> adj[205];
bool AM[205][205];
bitset<20010> isPrime;
vector<int> primes;
void sieve() {
  isPrime.set();
  isPrime[0] = isPrime[1] = 0;
  for (int i = 2; i <= MX; i++) {
    if (isPrime[i]) primes.push_back(i);
    for (int j = 0; j < primes.size() && i * primes[j] <= MX; j++) {
      isPrime[i * primes[j]] = 0;
      if (i % primes[j] == 0) break;
    }
  }
}
void prepgrf() {
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (isPrime[a[i] + a[j]]) {
        adj[i].push_back(j);
        adj[j].push_back(i);
        AM[i][j] = AM[j][i] = 1;
      }
    }
  }
  int odd = 0, even = 0;
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() == 1) CANNOT();
    if (a[i] & 1)
      odd++;
    else
      even++;
  }
  if (odd != even) CANNOT();
}
set<int> nxt[305];
struct Edge {
  int u, v;
  long long cap, flow;
  Edge() {}
  Edge(int u, int v, long long cap) : u(u), v(v), cap(cap), flow(0) {}
};
struct Dinic {
  int n;
  vector<Edge> E;
  vector<vector<int> > g;
  vector<int> d, pt;
  Dinic(int n) : n(n), E(0), g(n), d(n), pt(n) {}
  void AddEdge(int u, int v, long long cap) {
    E.emplace_back(u, v, cap);
    g[u].emplace_back(E.size() - 1);
    E.emplace_back(v, u, 0);
    g[v].emplace_back(E.size() - 1);
  }
  bool bfs(int s, int t) {
    fill(d.begin(), d.end(), n + 1);
    queue<int> q;
    q.push(s);
    d[s] = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      if (u == t) break;
      for (int k : g[u]) {
        Edge &e = E[k];
        if (e.flow < e.cap && d[e.v] > d[e.u] + 1) {
          d[e.v] = d[e.u] + 1;
          q.push(e.v);
        }
      }
    }
    return d[t] != n + 1;
  }
  long long dfs(int u, int t, long long flow = -1) {
    if (u == t || flow == 0) return flow;
    for (int &i = pt[u]; i < g[u].size(); i++) {
      Edge &e = E[g[u][i]];
      Edge &oe = E[g[u][i] ^ 1];
      if (d[e.v] == d[e.u] + 1) {
        long long amt = e.cap - e.flow;
        if (flow != -1 && amt > flow) amt = flow;
        if (long long pushed = dfs(e.v, t, amt)) {
          e.flow += pushed;
          oe.flow -= pushed;
          return pushed;
        }
      }
    }
    return 0;
  }
  long long MaxFlow(int s, int t) {
    long long ret = 0;
    while (bfs(s, t)) {
      fill(pt.begin(), pt.end(), 0);
      while (long long flow = dfs(s, t)) ret += flow;
    }
    return ret;
  }
};
int main() {
  sieve();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  prepgrf();
  Dinic dinic(n + 5);
  int S = 0, T = n + 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1) {
      dinic.AddEdge(S, i, 2);
      for (int nx : adj[i]) {
        dinic.AddEdge(i, nx, 1);
      }
    } else {
      dinic.AddEdge(i, T, 2);
    }
  }
  if (dinic.MaxFlow(S, T) != n) CANNOT();
  for (Edge &e : dinic.E) {
    if (e.u == S || e.u == T || e.v == S || e.v == T) continue;
    if ((a[e.u] & 1) && e.cap > 0 && e.flow == 1) {
      nxt[e.u].insert(e.v);
      nxt[e.v].insert(e.u);
    }
  }
  vector<vector<int> > ANS;
  vector<int> curr;
  for (int i = 1; i <= n; i++) {
    if (nxt[i].size()) {
      curr.clear();
      int u = i;
      curr.push_back(u);
      int v = *(nxt[u].begin());
      nxt[u].erase(v);
      nxt[v].erase(u);
      u = v;
      while (u != i) {
        curr.push_back(u);
        v = *(nxt[u].begin());
        nxt[u].erase(v);
        nxt[v].erase(u);
        u = v;
      }
      ANS.push_back(curr);
      curr.clear();
    }
  }
  printf("%lu\n", ANS.size());
  for (vector<int> &v : ANS) {
    int t = v.size();
    printf("%d", t);
    for (int i = 0; i < t; i++) printf(" %d", v[i]);
    puts("");
  }
  return 0;
}
