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;
const int maxn = 208, maxt = 208, Inf = 1e9;
int n, m, total, S, T, lay[maxt], a[maxn], cnt;
queue<int> Q;
bool p[100008], np[100008], vis[maxn];
vector<int> ans[maxn];
struct Bow {
  Bow *n, *r;
  int v, c;
  Bow() {}
} * h[maxt];
inline void AddBow(Bow *&x, int u, int v, int c) {
  x->v = v;
  x->c = c;
  x->n = h[u];
  h[u] = x;
}
inline void AddLine(int u, int v, int c) {
  Bow *x = new Bow();
  Bow *y = new Bow();
  AddBow(x, u, v, c);
  AddBow(y, v, u, 0);
  x->r = y;
  y->r = x;
}
bool Strat() {
  for (int i = 0; i < total; i++) lay[i] = -1;
  Q.push(S);
  lay[S] = 0;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (Bow *i = h[u]; i; i = i->n)
      if (lay[i->v] == -1 && i->c) Q.push(i->v), lay[i->v] = lay[u] + 1;
  }
  return lay[T] != -1;
}
int Augment(int u, int low) {
  int rep = 0;
  if (u == T) return low;
  for (Bow *i = h[u]; i && low; i = i->n)
    if (lay[i->v] == lay[u] + 1 && i->c) {
      int tmp = Augment(i->v, min(i->c, low));
      low -= tmp;
      i->c -= tmp;
      rep += tmp;
      i->r->c += tmp;
    }
  return rep;
}
inline int MaxFlow() {
  int rep = 0;
  while (Strat()) rep += Augment(S, Inf);
  return rep;
}
void Find(int u, int f, int c) {
  if (vis[u]) return;
  vis[u] = true;
  for (Bow *i = h[u]; i; i = i->n) {
    if (a[u] & 1) {
      if (i->c == 0 && i->v != f) {
        Find(i->v, u, c + 1);
        break;
      }
    } else {
      if (i->c == 1 && i->v != f) {
        Find(i->v, u, c + 1);
        break;
      }
    }
  }
  ans[cnt].push_back(u + 1);
}
int main() {
  scanf("%d", &n);
  total = n;
  S = total++;
  T = total++;
  for (int i = 2; i <= 20000; i++)
    if (!np[i]) {
      p[i] = true;
      for (int j = 2 * i; j <= 20000; j += i) np[j] = true;
    }
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    for (int j = 0; j < i; j++)
      if (p[a[i] + a[j]]) {
        if (a[i] & 1)
          AddLine(i, j, 1);
        else
          AddLine(j, i, 1);
      }
    if (a[i] & 1)
      AddLine(S, i, 2);
    else
      AddLine(i, T, 2);
  }
  int haha = MaxFlow();
  if (haha < n) {
    puts("Impossible");
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (!vis[i]) Find(i, 0, 0), cnt++;
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) {
    printf("%d ", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf("%d ", ans[i][j]);
    puts("");
  }
  return 0;
}