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;
int N;
int A[205];
int cnt[2];
bitset<20005> compo;
int cap[205][205];
int par[205];
int macflow;
vector<int> pp[205];
vector<int> adj[205];
void edge(int u, int v, int c) {
  cap[u][v] = c;
  adj[u].push_back(v);
  adj[v].push_back(u);
}
void match(int u, int v) {
  pp[u].push_back(v);
  pp[v].push_back(u);
}
void input() {
  scanf("%d", &N);
  for (int i = 2; i <= 20000; i++) {
    if (compo[i]) continue;
    for (int j = i * i; j <= 20000; j += i) {
      compo[j] = 1;
    }
  }
  for (int i = 1; i <= N; i++) {
    scanf("%d", &A[i]);
    if (A[i] & 1)
      edge(0, i, 2);
    else
      edge(i, 201, 2);
    for (int j = 1; j < i; j++) {
      if (compo[A[i] + A[j]] == 0) {
        if (A[i] & 1)
          edge(i, j, 1);
        else
          edge(j, i, 1);
      }
    }
  }
}
void DFS(int u) {
  if (u == 201) return;
  for (int v : adj[u]) {
    if (cap[u][v] == 0) continue;
    if (par[v] == -1) {
      par[v] = u;
      DFS(v);
    }
    if (par[201] != -1) return;
  }
}
int augment(int u, int c) {
  if (u == 0) return c;
  if (par[u] != -1) {
    int flow = augment(par[u], min(c, cap[par[u]][u]));
    cap[par[u]][u] -= flow;
    cap[u][par[u]] += flow;
    return flow;
  }
  return 0;
}
bool vis[205];
int pa;
vector<int> res[100];
void DFS2(int u, int p) {
  vis[u] = 1;
  res[pa].push_back(u);
  for (int v : pp[u]) {
    if (vis[v]) continue;
    DFS2(v, u);
  }
}
void ford() {
  while (1) {
    memset(par, -1, sizeof(par));
    par[0] = -2;
    DFS(0);
    int flow = augment(201, 1000000);
    if (flow == 0) break;
    macflow += flow;
  }
  if (macflow != N) {
    puts("Impossible");
    return;
  }
  for (int i = 1; i <= N; i++) {
    for (int j = i + 1; j <= N; j++) {
      if (compo[A[i] + A[j]] == 0) {
        if (A[i] & 1) {
          if (cap[j][i] == 1) {
            match(i, j);
          }
        } else {
          if (cap[i][j] == 1) {
            match(i, j);
          }
        }
      }
    }
  }
  for (int i = 1; i <= N; i++) {
    if (vis[i]) continue;
    pa++;
    DFS2(i, 0);
  }
  printf("%d\n", pa);
  for (int i = 1; i <= pa; i++) {
    printf("%d ", (int)res[i].size());
    for (int now : res[i]) {
      printf("%d ", now);
    }
    putchar('\n');
  }
  cin >> N;
}
int main() {
  input();
  ford();
}
