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;
int i, j, n, e = 0, o = 0, k;
int age[202], FLOW = 0;
int cap[202][202];
int flow[202][202];
int visit[202], por[202];
int prime[20002];
int tc = 0;
queue<int> q, t[202];
void genprime(int m) {
  prime[1] = -1;
  for (i = 2; i <= m; i++) {
    if (prime[i] == 0) {
      prime[i] = 1;
      for (j = 2; j * i <= m; j++) {
        prime[j * i] = -1;
      }
    }
  }
  return;
}
void bfs() {
  if (q.front() == n + 1) {
    FLOW++;
    k = n + 1;
    while (k != 0) {
      flow[por[k]][k]++;
      flow[k][por[k]]--;
      k = por[k];
    }
    for (i = 0; i <= n + 1; i++) {
      visit[i] = 0;
    }
    while (!q.empty()) {
      q.pop();
    }
    visit[0] = 1;
    q.push(0);
  } else {
    for (i = 0; i <= n + 1; i++) {
      if (cap[q.front()][i] - flow[q.front()][i] > 0 and visit[i] == 0) {
        por[i] = q.front();
        visit[i] = 1;
        q.push(i);
      }
    }
    q.pop();
  }
  return;
}
void dfs(int p) {
  visit[p] = 1;
  t[tc].push(p);
  for (j = 1; j <= n; j++) {
    if (flow[p][j] != 0 and visit[j] == 0) {
      dfs(j);
      break;
    }
  }
  return;
}
int main() {
  genprime(20000);
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &age[i]);
    if (age[i] % 2 == 0) {
      cap[0][i] = 2;
      e++;
    } else {
      cap[i][n + 1] = 2;
      o++;
    }
  }
  if (e != o) {
    printf("Impossible");
    return 0;
  }
  for (i = 1; i <= n; i++) {
    for (j = i + 1; j <= n; j++) {
      if (prime[age[i] + age[j]] == 1) {
        if (age[i] % 2 == 0) {
          cap[i][j] = 1;
        } else {
          cap[j][i] = 1;
        }
      }
    }
  }
  visit[0] = 1;
  q.push(0);
  while (!q.empty()) {
    bfs();
  }
  if (FLOW != n) {
    printf("Impossible");
    return 0;
  }
  for (i = 0; i <= n + 1; i++) {
    visit[i] = 0;
  }
  for (i = 1; i <= n; i++) {
    if (visit[i] == 0) {
      tc++;
      dfs(i);
    }
  }
  printf("%d\n", tc);
  for (i = 1; i <= tc; i++) {
    printf("%d ", t[i].size());
    while (!t[i].empty()) {
      printf("%d ", t[i].front());
      t[i].pop();
    }
    printf("\n");
  }
  return 0;
}