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 bian {
  int next, point, f;
} b[80000];
int a[300], n, totpoin, dst[300], pd[300], p[300], len, B[300][300], tot,
    x[300], totpoint, inf = 1e5;
vector<int> A[300];
void dfs(int k) {
  A[tot].push_back(k);
  pd[k] = 1;
  for (int i = 1; i <= n; i++)
    if (B[k][i] && pd[i] == 0) dfs(i);
}
void ade(int k1, int k2, int k3) {
  len++;
  b[len].point = k2;
  b[len].next = p[k1];
  b[len].f = k3;
  p[k1] = len;
}
void add(int k1, int k2, int k3) {
  ade(k1, k2, k3);
  ade(k2, k1, 0);
}
bool bfs() {
  int head = 1, now = 0, i, j;
  memset(dst, 0xff, sizeof dst);
  memset(pd, 0x00, sizeof pd);
  x[1] = 0;
  pd[0] = 1;
  dst[0] = 0;
  while (head > now) {
    now++;
    i = p[x[now]];
    while (i != -1) {
      j = b[i].point;
      if ((b[i].f) && (!pd[j])) {
        pd[j] = 1;
        dst[j] = dst[x[now]] + 1;
        if (j == totpoint) {
          return 1;
        }
        head++;
        x[head] = j;
      }
      i = b[i].next;
    }
  }
  return pd[totpoint];
}
int change(int k1, int k2) {
  if ((k1 == totpoint) || (k2 == 0)) return k2;
  int num = 0, k, i, j;
  i = p[k1];
  while (i != -1) {
    j = b[i].point;
    if ((b[i].f) && (dst[k1] + 1 == dst[j])) {
      k = change(j, min(k2, b[i].f));
      k2 = k2 - k;
      num += k;
      b[i].f = b[i].f - k;
      b[i ^ 1].f += k;
      if (k2 == 0) {
        break;
      }
    }
    i = b[i].next;
  }
  if (!num) {
    dst[k1] = -1;
  }
  return num;
}
int dinic() {
  int now = 0;
  while (bfs()) now += change(0, inf);
  return now;
}
int check(int k) {
  for (int i = 2; i * i <= k; i++)
    if (k % i == 0) return 0;
  return 1;
}
int main() {
  scanf("%d", &n);
  memset(p, 0xff, sizeof p);
  len = -1;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++)
    if (a[i] % 2)
      add(0, i, 2);
    else
      add(i, n + 1, 2);
  for (int i = 1; i <= n; i++)
    if (a[i] % 2)
      for (int j = 1; j <= n; j++)
        if ((a[j] % 2 == 0) && (check(a[i] + a[j]))) {
          add(i, j, 1);
        }
  totpoint = n + 1;
  if (dinic() != n) {
    printf("Impossible\n");
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (a[i] % 2)
      for (int j = p[i]; j != -1; j = b[j].next)
        if (j % 2 == 0 && b[j].f == 0) {
          B[i][b[j].point] = 1;
          B[b[j].point][i] = 1;
        }
  memset(pd, 0x00, sizeof pd);
  for (int i = 1; i <= n; i++)
    if (pd[i] == 0) {
      tot++;
      dfs(i);
    }
  cout << tot << endl;
  for (int i = 1; i <= tot; i++) {
    cout << A[i].size() << " ";
    for (int j = 0; j < A[i].size(); j++) cout << A[i][j] << " ";
    cout << endl;
  }
  return 0;
}