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 EDGE {
  int adj, w, next;
} edge[205 * 205 * 8];
int n, top, prime[30000], v[30000], gh[205], boot[205], X[205 * 205 * 8],
    Y[205 * 205 * 8], a[205], S, T;
int ans[205][205];
void addedge(int x, int y, int w) {
  edge[++top].adj = y;
  edge[top].w = w;
  edge[top].next = gh[x];
  gh[x] = top;
  edge[++top].adj = x;
  edge[top].w = 0;
  edge[top].next = gh[y];
  gh[y] = top;
}
int dist[205], q[205];
int bfs() {
  memset(dist, 0, sizeof(dist));
  int head = 0, tail = 1;
  q[1] = S;
  dist[S] = 1;
  while (head < tail) {
    int x = q[++head];
    for (int p = gh[x]; p; p = edge[p].next)
      if (edge[p].w && !dist[edge[p].adj]) {
        dist[edge[p].adj] = dist[x] + 1;
        q[++tail] = edge[p].adj;
      }
  }
  return dist[T];
}
int dinic(int x, int delta) {
  if (x == T) return delta;
  int dd = 0, tmp;
  for (int &p = boot[x]; p && delta; p = edge[p].next)
    if (edge[p].w && dist[edge[p].adj] == dist[x] + 1) {
      tmp = dinic(edge[p].adj, min(delta, edge[p].w));
      edge[p].w -= tmp;
      edge[p ^ 1].w += tmp;
      dd += tmp;
      delta -= tmp;
      if (dd) return dd;
    }
  if (!dd) dist[x] = -1;
  return dd;
}
bool Init() {
  top = 0;
  memset(v, 0, sizeof(v));
  for (int i = 2; i < 30000; i++) {
    if (!v[i]) prime[++top] = i;
    for (int j = 1; j <= top && prime[j] * i < 30000; j++) {
      v[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  scanf("%d", &n);
  int odd = 0, even = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] & 1)
      odd++;
    else
      even++;
  }
  if (odd != even) return false;
  top = 1;
  for (int i = 1; i <= n; i++)
    if (a[i] % 2 == 1)
      for (int j = 1; j <= n; j++)
        if ((a[j] % 2 == 0) && (v[a[i] + a[j]] == 0)) addedge(i, j, 1);
  S = n + 1;
  T = n + 2;
  for (int i = 1; i <= n; i++)
    if (a[i] & 1) addedge(S, i, 2);
  for (int i = 1; i <= n; i++)
    if (!(a[i] & 1)) addedge(i, T, 2);
  int pans = 0;
  while (bfs()) {
    memcpy(boot, gh, sizeof(gh));
    int F;
    while (F = dinic(S, 0x7fffffff)) pans += F;
  }
  return pans == n;
}
void work(int x) {
  v[x] = 1;
  X[++top] = x;
  for (int p = gh[x]; p; p = edge[p].next)
    if (!v[edge[p].adj]) {
      work(edge[p].adj);
      return;
    }
}
void Main() {
  int tt = 0;
  for (int p = 2; p <= top; p++)
    if ((p & 1) == 0 && edge[p].w == 0 && edge[p].adj != S &&
        edge[p].adj != T && edge[p ^ 1].adj != S && edge[p ^ 1].adj != T) {
      ++tt;
      X[tt] = edge[p].adj;
      Y[tt] = edge[p ^ 1].adj;
    }
  top = 0;
  memset(gh, 0, sizeof(gh));
  for (int i = 1; i <= tt; i++) addedge(X[i], Y[i], 0);
  memset(v, 0, sizeof(v));
  tt = 0;
  for (int i = 1; i <= n; i++)
    if (!v[i]) {
      top = 0;
      work(i);
      ++tt;
      ans[tt][0] = top;
      for (int j = 1; j <= top; j++) ans[tt][j] = X[j];
    }
  printf("%d\n", tt);
  for (int i = 1; i <= tt; i++) {
    top = ans[i][0];
    printf("%d", top);
    for (int j = 1; j <= top; j++) printf(" %d", ans[i][j]);
    puts("");
  }
}
int main() {
  if (!Init()) {
    puts("Impossible");
    return 0;
  }
  Main();
  return 0;
}