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;
vector<vector<int> > graph, Rgraph, conn, rrgraph;
vector<int> parent;
int n, s, t, V;
bool bfs() {
  bool visited[V];
  for (int i = 0; i < V; ++i) visited[i] = 0;
  queue<int> q;
  q.push(s);
  visited[s] = true;
  parent[s] = -1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int v = 0; v < V; v++) {
      if (visited[v] == false && Rgraph[u][v] > 0) {
        q.push(v);
        parent[v] = u;
        visited[v] = true;
      }
    }
  }
  return (visited[t] == true);
}
int fordFulkerson() {
  int u, v;
  parent.assign(V, -1);
  int max_flow = 0;
  while (bfs()) {
    int path_flow = INT_MAX;
    for (v = t; v != s; v = parent[v]) {
      u = parent[v];
      path_flow = min(path_flow, Rgraph[u][v]);
    }
    for (v = t; v != s; v = parent[v]) {
      u = parent[v];
      Rgraph[u][v] -= path_flow;
      Rgraph[v][u] += path_flow;
    }
    max_flow += path_flow;
  }
  return max_flow;
}
int limit = 20000;
vector<int> lp, prime;
void linear_sieve() {
  lp.assign(limit + 1, 0);
  for (int i = 2; i <= limit; ++i) {
    if (lp[i] == 0) {
      lp[i] = i;
      prime.push_back(i);
    }
    for (int j = 0;
         j < prime.size() && prime[j] <= lp[i] && i * prime[j] <= limit; ++j) {
      lp[i * prime[j]] = prime[j];
    }
  }
}
int main() {
  linear_sieve();
  scanf("%d", &n);
  int a[n], i, j, odd = 0, even = 0;
  V = n + 2;
  s = 0, t = V - 1;
  conn.assign(n, vector<int>(n, 0));
  graph.assign(V, vector<int>(V, 0));
  Rgraph.assign(V, vector<int>(V, 0));
  rrgraph.assign(V, vector<int>(V, 0));
  for (i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] % 2 == 0) ++even;
    if (a[i] % 2 == 1) ++odd;
  }
  for (i = 0; i < n; ++i) {
    for (j = 0; j < n; ++j) {
      if (lp[a[i] + a[j]] == a[i] + a[j]) {
        conn[i][j] = 1;
      }
    }
  }
  if (odd != even) {
    printf("Impossible");
    return 0;
  }
  for (i = 0; i < n; ++i) {
    if (a[i] % 2 == 0) {
      graph[i + 1][n + 1] = 2;
      Rgraph[i + 1][n + 1] = 2;
    }
    if (a[i] % 2 == 1) {
      graph[0][i + 1] = 2;
      Rgraph[0][i + 1] = 2;
    }
  }
  for (i = 0; i < n; ++i) {
    for (j = 0; j < n; ++j) {
      if (conn[i][j] == 1) {
        if (a[i] % 2 == 1) {
          graph[i + 1][j + 1] = 1;
          Rgraph[i + 1][j + 1] = 1;
        }
        if (a[i] % 2 == 0) {
          graph[j + 1][i + 1] = 1;
          Rgraph[j + 1][i + 1] = 1;
        }
      }
    }
  }
  int ans = fordFulkerson();
  for (i = 1; i < V - 1; ++i) {
    for (j = 1; j < V - 1; ++j) {
      rrgraph[j][i] = Rgraph[i][j];
    }
  }
  vector<int> chose(n + 1, 0);
  vector<vector<int> > res;
  int k;
  for (k = 1; k <= n; ++k) {
    if (chose[k] == 0) {
      i = k;
      res.push_back(vector<int>());
      int begin = i;
      res[res.size() - 1].push_back(i);
      do {
        int check = 0;
        chose[i] = 1;
        if (a[i - 1] % 2 == 1) {
          for (j = 1; j < V - 1; ++j) {
            if (rrgraph[i][j] != 0 && chose[j] == 0) {
              res[res.size() - 1].push_back(j);
              i = j;
              check = 1;
              break;
            }
          }
        } else {
          for (j = 1; j < V - 1; ++j) {
            if (Rgraph[i][j] != 0 && chose[j] == 0) {
              res[res.size() - 1].push_back(j);
              i = j;
              check = 1;
              break;
            }
          }
        }
        if (check == 0) break;
      } while (1);
    }
  }
  for (i = 0; i < res.size(); ++i) {
    if (res[i].size() < 3) {
      printf("Impossible");
      return 0;
    }
  }
  printf("%d\n", res.size());
  for (i = 0; i < res.size(); ++i) {
    printf("%d ", res[i].size());
    for (j = 0; j < res[i].size(); ++j) {
      printf("%d ", res[i][j]);
    }
    printf("\n");
  }
}