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 = 200 + 5, inf = 100;
typedef int ary[maxn];
bitset<20000> ntp;
int s = 0, t, n, m = 1, tot = 0, prm[2600], stream = 0;
ary que, numb, a, head, neck;
struct node {
  int don, nxt, rst;
  void ins(int x, int y, int f) {
    don = y;
    rst = f;
    nxt = head[x];
    head[x] = m;
  }
} star[maxn * maxn];
bool BFS() {
  memset(numb, 0xff, sizeof(numb));
  numb[que[0] = s] = 0;
  for (int f = 0, r = 0, now = que[0]; f <= r; now = que[++f]) {
    neck[now] = head[now];
    for (int q = head[now]; q; q = star[q].nxt)
      if (numb[star[q].don] < 0 && star[q].rst) {
        numb[star[q].don] = numb[now] + 1;
        que[++r] = star[q].don;
      }
  }
  return numb[t] < 0 ? 0 : 1;
}
int DFS(int now, int lim) {
  if (now == t) return lim;
  int used = 0;
  for (int q = neck[now]; q && lim > used || (neck[now] = q, 0);
       q = star[q].nxt)
    if (numb[now] + 1 == numb[star[q].don] && star[q].rst) {
      int flow = DFS(star[q].don, min(lim - used, star[q].rst));
      used += flow;
      star[q].rst -= flow;
      star[q ^ 1].rst += flow;
    }
  return used;
}
void prepare() {
  for (int i = 2; i < 20000; i++) {
    if (!ntp[i]) prm[++tot] = i;
    for (int j = 1; j <= tot && i * prm[j] < 20000; j++) {
      ntp[i * prm[j]] = 1;
      if (i % prm[j] == 0) break;
    }
  }
  scanf("%d", &n);
  t = n + 1;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      if (!ntp[a[i] + a[j]]) {
        if (a[i] & 1)
          star[++m].ins(i, j, 1), star[++m].ins(j, i, 0);
        else
          star[++m].ins(j, i, 1), star[++m].ins(i, j, 0);
      }
  for (int i = 1; i <= n; i++)
    if (a[i] & 1)
      star[++m].ins(s, i, 2), star[++m].ins(i, s, 0);
    else
      star[++m].ins(i, t, 2), star[++m].ins(t, i, 0);
}
void cir(int now, int num, int ord) {
  numb[now] = num * maxn * maxn + ord * maxn + now;
  for (int q = head[now]; q; q = star[q].nxt)
    if (!numb[star[q].don] &&
        (a[now] & 1 && !star[q].rst || !(a[now] & 1) && star[q].rst))
      cir(star[q].don, num, ord + 1);
}
bool cmp(int a, int b) { return a > b ? 1 : 0; }
int main() {
  prepare();
  while (BFS()) stream += DFS(s, inf);
  if (stream < n) {
    printf("Impossible\n");
    return 0;
  }
  memset(numb, 0, sizeof(numb));
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (!numb[i]) cir(i, ++cnt, 1);
  printf("%d\n", cnt);
  sort(numb + 1, numb + n + 1, cmp);
  for (int i = cnt, j = 1; i; i--) {
    printf("%d", numb[j] / maxn % maxn);
    for (; j <= n && numb[j] / maxn / maxn == i; j++)
      printf(" %d", numb[j] % maxn);
    printf("\n");
  }
  return 0;
}