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 N = 200 + 5;
bool sito[20005];
void era() {
  for (int i = 2; i <= 20000; i++)
    if (sito[i] == 0)
      for (int j = i * i; j <= 20000; j += i) sito[j] = 1;
}
int tab[N], lef[N], rig[N], n;
bool g[N][N], vis[2][N];
bool dfs(int x, int co) {
  if (vis[co][x]) return 0;
  vis[co][x] = 1;
  for (int i = 1; i <= n; i++)
    if (g[x][i]) {
      if (co == 1 && i != lef[x])
        if (lef[i] == 0 || dfs(lef[i], 1)) {
          rig[x] = i;
          lef[i] = x;
          return 1;
        }
      if (co == 0 && i != rig[x])
        if (rig[i] == 0 || dfs(rig[i], 0)) {
          lef[x] = i;
          rig[i] = x;
          return 1;
        }
    }
  return 0;
}
vector<vector<int> > v;
vector<int> pusty;
void przejdz(int x) {
  if (vis[0][x]) return;
  v.back().push_back(x);
  vis[0][x] = 1;
  przejdz(lef[x]);
}
int main() {
  era();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", tab + i);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (sito[tab[i] + tab[j]] == 0) {
        g[i][j] = 1;
      }
  bool OK = 1;
  while (OK) {
    OK = 0;
    for (int i = 1; i <= n; i++) vis[0][i] = vis[1][i] = 0;
    for (int i = 1; i <= n; i++) {
      if (rig[i] == 0)
        if (dfs(i, 1)) OK = 1;
      if (lef[i] == 0)
        if (dfs(i, 0)) OK = 1;
    }
  }
  OK = 1;
  for (int i = 1; i <= n; i++)
    if (rig[i] == 0 || lef[i] == 0) OK = 0;
  if (OK == 0)
    printf("Impossible\n");
  else {
    for (int i = 1; i <= n; i++) vis[0][i] = 0;
    for (int i = 1; i <= n; i++)
      if (vis[0][i] == 0) {
        v.push_back(pusty);
        przejdz(i);
      }
    printf("%d\n", v.size());
    for (int i = 0; i < v.size(); i++) {
      printf("%d ", v[i].size());
      for (int j = 0; j < v[i].size(); j++) printf("%d ", v[i][j]);
      printf("\n");
    }
  }
}