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 node {
  int v, flow, next;
} edge[40010];
int n, Prime[20010], num[210], Head[210], d[210], e, desk[210][210], vis[210],
    INF = 1e9, link[210][210];
queue<int> qu;
void add(int u, int v, int f) {
  edge[e].v = v;
  edge[e].flow = f;
  edge[e].next = Head[u];
  Head[u] = e;
  e++;
}
void init() {
  int i, j, k;
  e = 0;
  memset(Head, -1, sizeof(Head));
  for (i = 1; i <= n; i++)
    if (num[i] & 1) {
      add(0, i, 2);
      add(i, 0, 2);
    } else {
      add(i, n + 1, 2);
      add(n + 1, i, 0);
    }
  for (i = 1; i <= n; i++)
    for (j = i + 1; j <= n; j++)
      if (Prime[num[i] + num[j]] == 0) {
        if (num[i] & 1) {
          add(i, j, 1);
          add(j, i, 0);
        } else {
          add(j, i, 1);
          add(i, j, 0);
        }
      }
}
int bfs() {
  int i, j, u, v;
  memset(d, -1, sizeof(d));
  while (!qu.empty()) qu.pop();
  d[0] = 0;
  qu.push(0);
  while (!qu.empty()) {
    u = qu.front();
    qu.pop();
    for (j = Head[u]; j != -1; j = edge[j].next)
      if (edge[j].flow > 0 && d[edge[j].v] == -1) {
        d[edge[j].v] = d[u] + 1;
        if (edge[j].v == n + 1) return 1;
        qu.push(edge[j].v);
      }
  }
  return 0;
}
int dfs(int u, int f) {
  if (u == n + 1 || f == 0) return f;
  int ans = 0, v, i, j, k;
  for (j = Head[u]; j != -1; j = edge[j].next)
    if (edge[j].flow > 0 && d[edge[j].v] == d[u] + 1) {
      k = dfs(edge[j].v, min(edge[j].flow, f));
      edge[j].flow -= k;
      edge[j ^ 1].flow += k;
      ans += k;
      f -= k;
      if (f == 0) break;
    }
  d[u] = -1;
  return ans;
}
int dinic() {
  int ans = 0;
  while (bfs()) ans += dfs(0, INF);
  return ans;
}
void dfs2(int p, int u) {
  int i, j, k;
  for (i = 1; i <= n; i++) {
    if (vis[i] == 0 && link[u][i] == 1) {
      vis[i] = 1;
      desk[p][0]++;
      desk[p][desk[p][0]] = i;
      dfs2(p, i);
    }
  }
}
int main() {
  int i, j, k, ans, p;
  for (i = 2; i <= 200; i++)
    if (Prime[i] == 0)
      for (j = i * 2; j <= 20000; j += i) Prime[j] = 1;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &num[i]);
  init();
  ans = dinic();
  if (ans != n)
    printf("Impossible\n");
  else {
    for (i = 1; i <= n; i++)
      if (num[i] & 1)
        for (j = Head[i]; j != -1; j = edge[j].next)
          if (edge[j].flow == 0) {
            link[i][edge[j].v] = 1;
            link[edge[j].v][i] = 1;
          }
    p = 0;
    for (i = 1; i <= n; i++)
      if (vis[i] == 0) {
        vis[i] = 1;
        p++;
        desk[p][0] = 1;
        desk[p][1] = i;
        dfs2(p, i);
      }
    printf("%d\n", p);
    for (i = 1; i <= p; i++) {
      printf("%d", desk[i][0]);
      for (j = 1; j <= desk[i][0]; j++) printf(" %d", desk[i][j]);
      printf("\n");
    }
  }
}