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 = 210, maxm = 40010, inf = 1e8;
int n, S, T, m, tot, a[maxn], lnk[maxn], son[maxm], nxt[maxm], que[maxn],
    w[maxm], dep[maxn], cur[maxn], fa[maxn], ans[maxn][maxn];
bool check(int x) {
  int sq = sqrt(x);
  for (int i = 2; i <= sq; i++)
    if (x % i == 0) return 0;
  return 1;
}
void add(int x, int y, int z) {
  son[++tot] = y, w[tot] = z, nxt[tot] = lnk[x], lnk[x] = tot;
  son[++tot] = x, w[tot] = 0, nxt[tot] = lnk[y], lnk[y] = tot;
}
bool bfs() {
  int head = 0, tail = 1;
  que[tail] = S;
  memset(dep, 255, sizeof(dep));
  dep[S] = 0;
  while (head < tail) {
    head++;
    for (int j = lnk[que[head]]; j != -1; j = nxt[j])
      if (dep[son[j]] == -1 && w[j] > 0) {
        dep[son[j]] = dep[que[head]] + 1;
        que[++tail] = son[j];
      }
  }
  if (dep[T] == -1)
    return 0;
  else
    return 1;
}
int dfs(int x, int flow) {
  if (x == T) return flow;
  for (int& j = cur[x]; j != -1; j = nxt[j])
    if (dep[son[j]] == dep[x] + 1 && w[j] != 0) {
      int x = dfs(son[j], min(flow, w[j]));
      if (x) {
        w[j] -= x;
        w[j ^ 1] += x;
        return x;
      }
    }
  return 0;
}
int dinic() {
  int ans = 0;
  while (bfs()) {
    for (int i = 1; i <= T; i++) cur[i] = lnk[i];
    int sum;
    while (sum = dfs(S, inf)) ans += sum;
  }
  return ans;
}
void find(int x, int sum) {
  fa[x] = sum;
  ans[sum][++ans[sum][0]] = x;
  if (a[x] % 2 == 1) {
    for (int j = lnk[x]; j != -1; j = nxt[j])
      if (w[j] == 0 && !fa[son[j]]) find(son[j], sum);
  } else {
    for (int j = lnk[x]; j != -1; j = nxt[j])
      if (w[j ^ 1] == 0 && !fa[son[j]]) find(son[j], sum);
  }
}
int main() {
  tot = -1;
  memset(lnk, 255, sizeof(lnk));
  memset(nxt, 255, sizeof(nxt));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  T = n + 2, S = n + 1;
  for (int i = 1; i <= n; i++)
    if (a[i] % 2 == 1)
      add(S, i, 2);
    else
      add(i, T, 2);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i] % 2 == 1 && a[j] % 2 == 0 && check(a[j] + a[i])) add(i, j, 1);
  int flo = dinic();
  if (flo != n) {
    printf("Impossible\n");
    return 0;
  }
  int s = 0;
  for (int i = 1; i <= n; i++)
    if (!fa[i]) find(i, ++s);
  printf("%d\n", s);
  for (int i = 1; i <= s; i++) {
    printf("%d ", ans[i][0]);
    for (int j = 1; j <= ans[i][0]; j++) printf("%d ", ans[i][j]);
    printf("\n");
  }
  return 0;
}