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;
const long long M = 998244353;
int n, i, j, m, a[100005], head[500005], Next[1000005], adj[1000005],
    f[1000005], k = 1, flow[1000005];
int d[500005], q[500005], l, r, vis[205], ans, last[10005], fans, p[205][205],
    num[205], t;
void Push(int u, int v, int w) {
  Next[++k] = head[u];
  head[u] = k;
  adj[k] = v;
  flow[k] = w;
}
bool dfs(int i, int f) {
  int j;
  if (i == n + 1) {
    fans = f;
    return true;
  }
  for (j = last[i]; j != 0; last[i] = j = Next[j])
    if (d[adj[j]] == d[i] + 1 && flow[j] != 0)
      if (dfs(adj[j], min(f, flow[j]))) {
        flow[j] -= fans;
        flow[j ^ 1] += fans;
        return true;
      }
  return false;
}
void dfs2(int i) {
  int j;
  if (vis[i] == 1) return;
  vis[i] = 1;
  p[t][++num[t]] = i;
  for (j = head[i]; j != 0; j = Next[j])
    if (a[i] % 2) {
      if (flow[j] == 0) dfs2(adj[j]);
    } else {
      if (flow[j] == 1) dfs2(adj[j]);
    }
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 2; i * i <= 100000; i++)
    if (f[i] == 0)
      for (j = i * 2; j <= 100000; j += i) f[j] = 1;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++)
      if (f[a[i] + a[j]] == 0 && a[i] % 2) {
        Push(i, j, 1);
        Push(j, i, 0);
      }
  for (i = 1; i <= n; i++)
    if (a[i] % 2) {
      Push(0, i, 2);
      Push(i, 0, 0);
    } else {
      Push(i, n + 1, 2);
      Push(n + 1, i, 0);
    }
  while (1) {
    q[l = r = 1] = 0;
    memset(vis, 0, sizeof(vis));
    vis[0] = 1;
    while (l <= r) {
      for (i = head[q[l]]; i != 0; i = Next[i])
        if (vis[adj[i]] == 0 && flow[i] != 0) {
          vis[adj[i]] = 1;
          d[adj[i]] = d[q[l]] + 1;
          q[++r] = adj[i];
        }
      l++;
    }
    if (vis[n + 1] == 0) break;
    for (i = 0; i <= n + 1; i++) last[i] = head[i];
    while (dfs(0, 1 << 30)) ans += fans;
  }
  if (ans != n) {
    printf("Impossible");
    return 0;
  }
  memset(vis, 0, sizeof(vis));
  for (i = 1; i <= n; i++)
    if (vis[i] == 0) {
      t++;
      dfs2(i);
    }
  printf("%d\n", t);
  for (i = 1; i <= t; i++) {
    printf("%d ", num[i]);
    for (j = 1; j <= num[i]; j++) printf("%d ", p[i][j]);
    printf("\n");
  }
}
