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 n, k;
int l[250];
int r[250];
int rid[250];
int lid[250];
int pr[100500];
int lc, rc;
int cnt1, cnt2;
int nn[250];
int py[250];
int py2[250];
int px2[250];
vector<int> g[250];
vector<int> ans[250];
int c;
bool rec(int v) {
  if (nn[v]) return false;
  nn[v] = true;
  for (int i = 0; i < g[v].size(); ++i) {
    int x = g[v][i];
    if (py[x] == -1 || rec(py[x])) {
      py[x] = v;
      return true;
    }
  }
  return false;
}
bool rec2(int v) {
  if (nn[v]) return false;
  nn[v] = true;
  for (int i = 0; i < g[v].size(); ++i) {
    int x = g[v][i];
    if (x == -1) continue;
    if (py2[x] == -1 || rec2(py2[x])) {
      py2[x] = v;
      return true;
    }
  }
  return false;
}
void go(int v, int c) {
  if (nn[v]) return;
  nn[v] = true;
  ans[c].push_back(rid[v]);
  v = py[v];
  ans[c].push_back(lid[v]);
  v = px2[v];
  go(v, c);
}
int main() {
  for (int i = 2; i < 100500; ++i) pr[i] = true;
  for (int i = 2; i < 100500; ++i) {
    for (int j = i + i; j < 100500; j += i) pr[j] = false;
  }
  scanf("%i", &n);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%i", &x);
    if (x % 2 == 0) {
      l[lc] = x;
      lid[lc] = i;
      lc++;
    } else {
      r[rc] = x;
      rid[rc] = i;
      rc++;
    }
  }
  if (lc != rc) {
    printf("Impossible\n");
    return 0;
  }
  k = lc;
  for (int i = 0; i < k; ++i)
    for (int j = 0; j < k; ++j)
      if (pr[l[i] + r[j]]) g[i].push_back(j);
  for (int i = 0; i < k; ++i) py[i] = -1;
  for (int i = 0; i < k; ++i) {
    memset(nn, 0, sizeof(nn));
    if (rec(i)) cnt1++;
  }
  if (cnt1 != k) {
    printf("Impossible\n");
    return 0;
  }
  for (int i = 0; i < k; ++i) {
    int x = py[i];
    for (int j = 0; j < g[x].size(); ++j)
      if (g[x][j] == i) g[x][j] = -1;
  }
  for (int i = 0; i < k; ++i) py2[i] = -1;
  for (int i = 0; i < k; ++i) {
    memset(nn, 0, sizeof(nn));
    if (rec2(i)) cnt2++;
  }
  if (cnt2 != k) {
    printf("Impossible\n");
    return 0;
  }
  for (int i = 0; i < k; ++i) {
    px2[py2[i]] = i;
  }
  memset(nn, 0, sizeof(nn));
  for (int i = 0; i < k; ++i)
    if (!nn[i]) {
      go(i, c);
      c++;
    }
  printf("%i\n", c);
  for (int i = 0; i < c; ++i) {
    printf("%i", ans[i].size());
    for (int j = 0; j < ans[i].size(); ++j) printf(" %i", ans[i][j] + 1);
    printf("\n");
  }
  return 0;
}