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 inft = 1000000009;
const int MAXM = 1000006;
const int MAXN = 505;
vector<int> V[MAXN];
int c, d;
int M[MAXN];
int MT[MAXN];
bool odw[MAXN];
queue<int> Q;
inline bool dfs(int u) {
  odw[u] = 1;
  Q.push(u);
  for (typeof((V[u]).begin()) it = (V[u]).begin(); it != (V[u]).end(); ++it)
    if (MT[u] != *it)
      if ((M[*it] == -1) || (!odw[M[*it]] && dfs(M[*it]))) {
        M[u] = *it;
        M[*it] = u;
        return 1;
      }
  return 0;
}
int matching() {
  for (int i = 0; i < (c); ++i) odw[i] = 0;
  for (int i = 0; i < (c + d); ++i) M[i] = -1;
  bool czy = 1;
  while (czy) {
    for (czy = 0; !Q.empty(); Q.pop()) odw[Q.front()] = 0;
    for (int i = 0; i < (c); ++i)
      if (M[i] == -1 && !odw[i]) czy |= dfs(i);
  }
  int ret = 0;
  for (int i = 0; i < (c); ++i)
    if (M[i] != -1) ++ret;
  return ret;
}
vector<pair<int, int> > A, B;
bool prime[MAXM];
void solve() {
  for (int i = 0; i < (MAXM); ++i)
    if (i > 1)
      if (!prime[i]) {
        for (int j = 2 * i; j < MAXM; j += i) prime[j] = true;
      }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    int a;
    scanf("%d", &a);
    if (a % 2)
      A.push_back(pair<int, int>(a, i));
    else
      B.push_back(pair<int, int>(a, i));
  }
  c = A.size();
  d = B.size();
  for (int i = 0; i < (c); ++i)
    for (int j = 0; j < (d); ++j)
      if (!prime[A[i].first + B[j].first]) {
        V[i].push_back(c + j);
      }
  for (int i = 0; i < (n); ++i) MT[i] = -1;
  int ans = matching();
  for (int i = 0; i < (n); ++i) MT[i] = M[i];
  if (2 * ans != n) {
    printf("Impossible\n");
    return;
  }
  ans = matching();
  if (2 * ans != n) {
    printf("Impossible\n");
    return;
  }
  vector<int> C[MAXN];
  bool vis[MAXN];
  for (int i = 0; i < (MAXN); ++i) vis[i] = false;
  int nr = 0;
  for (int i = 0; i < (c); ++i)
    if (!vis[i]) {
      int tam = M[i], tu = MT[tam];
      vis[i] = true;
      vis[tu] = true;
      C[nr].push_back(i);
      C[nr].push_back(tam);
      C[nr].push_back(tu);
      while (tu != i) {
        tam = M[tu];
        tu = MT[tam];
        vis[i] = true;
        vis[tu] = true;
        C[nr].push_back(tam);
        C[nr].push_back(tu);
      }
      C[nr].pop_back();
      nr++;
    }
  printf("%d\n", nr);
  for (int i = 0; i < (nr); ++i) {
    printf("%d ", C[i].size());
    for (typeof((C[i]).begin()) it = (C[i]).begin(); it != (C[i]).end(); ++it) {
      if (*it < c)
        printf("%d ", A[*it].second + 1);
      else
        printf("%d ", B[*it - c].second + 1);
    }
    puts("");
  }
}
int main() {
  int t = 1;
  for (int i = 0; i < (t); ++i) solve();
  return 0;
}