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;
inline void __CIN__() {
  ios::sync_with_stdio(false);
  cin.tie(0);
}
char buf;
inline int xint() {
  while (buf = getchar(), buf < '0' || buf > '9')
    ;
  int x = buf - '0';
  for (; buf = getchar(), buf >= '0' && buf <= '9'; x = x * 10 + buf - '0')
    ;
  return x;
}
inline long long xll() {
  while (buf = getchar(), buf < '0' || buf > '9')
    ;
  long long x = buf - '0';
  for (; buf = getchar(), buf >= '0' && buf <= '9'; x = x * 10 + buf - '0')
    ;
  return x;
}
inline string xstring() {
  while (buf = getchar(), buf == ' ' || buf == '\n')
    ;
  string x = "";
  for (x += buf; buf = getchar(), buf != ' ' && buf != '\n'; x += buf)
    ;
  return x;
}
int n, data[205], pre[205];
vector<int> edge[205];
bool vis[205];
bool isPrime(int x) {
  for (int i = 2; i * i <= x; ++i)
    if (x % i == 0) return 0;
  return 1;
}
bool match(int now) {
  for (auto nxt : edge[now]) {
    if (!vis[nxt] && pre[now] != nxt) {
      vis[nxt] = 1;
      if (pre[nxt] == -1 || match(pre[nxt])) {
        pre[nxt] = now;
        return 1;
      }
    }
  }
  return 0;
}
int main() {
  n = xint();
  int i, j;
  for (i = 0; i < n; ++i) {
    data[i] = xint();
    for (j = 0; j < i; ++j)
      if (isPrime(data[i] + data[j])) {
        edge[i].push_back(j);
        edge[j].push_back(i);
      }
  }
  memset(pre, -1, sizeof(pre));
  for (i = 0; i < n; ++i) {
    memset(vis, 0, sizeof(vis));
    if (!match(i)) {
      printf("Impossible\n");
      return 0;
    }
  }
  vector<vector<int> > ans;
  int nxt[205];
  memset(nxt, -1, sizeof(nxt));
  for (i = 0; i < n; ++i) nxt[pre[i]] = i;
  while (1) {
    int pos = -1;
    for (i = 0; i < n; ++i)
      if (nxt[i] != -1) {
        pos = i;
        break;
      }
    if (pos == -1) {
      break;
    }
    vector<int> ansQ;
    while (nxt[pos] != -1) {
      ansQ.push_back(pos);
      int tmp = pos;
      pos = nxt[pos];
      nxt[tmp] = -1;
    }
    ans.push_back(ansQ);
  }
  printf("%d\n", ans.size());
  for (i = 0; i < ans.size(); ++i) {
    printf("%d", ans[i].size());
    for (auto x : ans[i]) printf(" %d", 1 + x);
    printf("\n");
  }
  return 0;
}