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;
struct ed {
  int to, c, f;
} edge[101010];
int n, seen[303], tempo, curr, s, t, comp[303], sz[303], vet[303], cont;
vector<int> adj[303], g[303], answer[303];
int init() {
  for (int i = 0; i <= n; i++) {
    comp[i] = i;
    sz[i] = 1;
  }
}
int find(int i) { return i == comp[i] ? i : comp[i] = find(comp[i]); }
bool same(int i, int j) { return find(i) == find(j); }
void join(int i, int j) {
  int x = find(i), y = find(j);
  if (x == y) return;
  sz[y] += sz[x];
  sz[x] = 0;
  comp[x] = y;
}
void add_edge(int a, int b, int c, int r) {
  edge[curr].to = b;
  edge[curr].c = c;
  edge[curr].f = 0;
  adj[a].push_back(curr++);
  edge[curr].to = a;
  edge[curr].c = r;
  edge[curr].f = 0;
  adj[b].push_back(curr++);
}
bool is_prime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
void build_graph() {
  curr = s = 0;
  t = 303 - 2;
  for (int i = 1; i <= n; i++) {
    if (vet[i] % 2)
      add_edge(s, i, 2, 0);
    else
      add_edge(i, t, 2, 0);
  }
  for (int i = 1; i <= n; i++) {
    if (!(vet[i] % 2)) continue;
    for (int j = 1; j <= n; j++) {
      if (j == i) continue;
      if (is_prime(vet[i] + vet[j])) add_edge(i, j, 1, 0);
    }
  }
}
int dfs(int u, int f) {
  if (u == t) return f;
  seen[u] = tempo;
  for (int i = 0; i < adj[u].size(); i++) {
    int e = adj[u][i];
    int v = edge[e].to;
    int w = edge[e].c - edge[e].f;
    if (seen[v] < tempo && w > 0) {
      if (int a = dfs(v, min(f, w))) {
        edge[e].f += a;
        edge[e ^ 1].f -= a;
        return a;
      }
    }
  }
  return 0;
}
int ford_fulk() {
  tempo = 1;
  memset(seen, 0, sizeof seen);
  int flow = 0;
  while (int a = dfs(s, 10101010)) {
    flow += a;
    tempo++;
  }
  return flow;
}
void dfs2(int u, int pai) {
  if (seen[u]) {
    return;
  }
  seen[u] = 1;
  answer[cont].push_back(u);
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == pai) continue;
    dfs2(v, u);
  }
}
void solve() {
  init();
  for (int i = 1; i <= n; i++) {
    if (!(vet[i] % 2)) continue;
    for (int j = 0; j < adj[i].size(); j++) {
      int e = adj[i][j];
      if (e % 2) continue;
      if (edge[e].f) {
        g[i].push_back(edge[e].to);
        g[edge[e].to].push_back(i);
        join(i, edge[e].to);
      }
    }
  }
  memset(seen, 0, sizeof seen);
  cont = 0;
  for (int i = 1; i <= n; i++) {
    if (!seen[i]) {
      answer[cont].push_back(sz[find(i)]);
      dfs2(i, -1);
      cont++;
    }
  }
  printf("%d\n", cont);
  for (int i = 0; i < cont; i++) {
    printf("%d", answer[i][0]);
    for (int j = 1; j < answer[i].size(); j++) printf(" %d", answer[i][j]);
    printf("\n");
  }
}
int main() {
  int par, imp = par = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> vet[i];
    if (vet[i] % 2)
      imp++;
    else
      par++;
  }
  if (par != imp) {
    printf("Impossible\n");
    return 0;
  }
  build_graph();
  if (ford_fulk() != n)
    printf("Impossible\n");
  else {
    solve();
  }
}