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;
int p[40010] = {0};
int adj[1001][1001];
int parent[2100];
vector<int> v[501];
int sz = 0;
int vis[501], a[501];
int bfs(int graph[][1001], int v, int t, int s) {
  int visited[1001] = {0}, i, u;
  queue<int> q;
  q.push(s);
  visited[s] = 1;
  for (i = 0; i <= v; i++) parent[i] = -1;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (i = 0; i < v; i++) {
      if (graph[u][i] > 0 && visited[i] == 0) {
        q.push(i);
        parent[i] = u;
        visited[i] = 1;
      }
    }
  }
  return visited[t];
}
int maxflow(int graph[][1001], int s, int t, int V) {
  int maxflow = 0, pathflow, v;
  while (bfs(graph, V, t, s)) {
    pathflow = 1000000000;
    for (v = t; v != s; v = parent[v])
      pathflow = min(pathflow, graph[parent[v]][v]);
    maxflow += pathflow;
    for (v = t; v != s; v = parent[v]) {
      graph[v][parent[v]] += pathflow;
      graph[parent[v]][v] -= pathflow;
    }
  }
  return maxflow;
}
void dfs(int u, int n) {
  if (u > n) {
    for (int i = 1; i <= n; i++)
      if (adj[u][i] > 0 && !vis[i]) {
        vis[i] = 1;
        v[sz].push_back(i);
        dfs(i, n);
      }
  } else {
    for (int i = n + 1; i <= 2 * n; i++)
      if (adj[i][u] > 0 && !vis[i]) {
        vis[i] = 1;
        v[sz].push_back(i - n);
        dfs(i, n);
      }
  }
}
int main() {
  int n, i, j, e = 0, o = 0;
  for (i = 2; i <= 200; i++)
    if (p[i] == 0)
      for (j = i + i; j <= 40000; j += i) p[j] = 1;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] % 2 == 0)
      e++;
    else
      o++;
  }
  if (e != o) {
    printf("Impossible\n");
    return 0;
  }
  for (i = 0; i <= 2 * n + 1; i++)
    for (j = 0; j <= 2 * n + 1; j++) adj[i][j] = 0;
  for (i = 0; i < n; i++)
    for (j = i + 1; j < n; j++)
      if (p[a[i] + a[j]] == 0) {
        if (a[i] % 2 == 1)
          adj[i + 1][j + n + 1] = 1;
        else
          adj[j + 1][i + n + 1] = 1;
      }
  for (i = 1; i <= n; i++)
    if (a[i - 1] % 2 == 1) adj[0][i] = 2;
  for (i = 1; i <= n; i++)
    if (a[i - 1] % 2 == 0) adj[i + n][2 * n + 1] = 2;
  int ans = maxflow(adj, 0, 2 * n + 1, 2 * n + 2);
  if (ans != n) {
    printf("Impossible\n");
    return 0;
  }
  sz = 0;
  for (i = n + 1; i <= 2 * n; i++) {
    if (!vis[i] && a[i - n - 1] % 2 == 0) {
      vis[i] = 1;
      v[sz].push_back(i - n);
      dfs(i, n);
      sz++;
    }
  }
  assert(sz <= n / 3);
  printf("%d\n", sz);
  for (i = 0; i < sz; i++) {
    printf("%d ", v[i].size());
    for (j = 0; j < v[i].size(); j++) printf("%d ", v[i][j]);
    printf("\n");
  }
  return 0;
}