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;
class pr {
 public:
  long long f[202][202], c[202][202], x[202];
  int h[202], s = 0, t = 202 - 1;
  vector<int> a[202];
  queue<int> q;
  pr() {
    memset(f, 0, sizeof f);
    memset(c, 0, sizeof c);
    memset(x, 0, sizeof x);
    memset(h, 0, sizeof h);
    h[s] = 202;
  }
  void send(int u, int v, long long z) {
    x[u] -= z;
    x[v] += z;
    f[u][v] += z;
    f[v][u] -= z;
    if (v != s and v != t) q.push(v);
  }
  bool push(int u, int v) {
    long long z = c[u][v] - f[u][v];
    if (x[u] > 0 and h[u] > h[v] and z > 0) {
      send(u, v, min(z, x[u]));
      return 1;
    }
    return 0;
  }
  void add(int u, int v, long long z, bool d = 0) {
    c[u][v] += z;
    a[u].push_back(v);
    if (u == s) send(u, v, z);
    if (d)
      a[v].push_back(u);
    else
      add(v, u, z, -1);
  }
  void calc() {
    while (!q.empty()) {
      int u = q.front();
      bool p = 0;
      for (int v : a[u])
        if (push(u, v)) p = 1;
      if (!p) h[u]++;
      if (x[u] == 0) q.pop();
    }
  }
} mf;
bool isPrime(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int n, cc = 0, vs[202];
long long a[202];
vector<int> ans[202];
void dfs(int p) {
  vs[p] = 1;
  ans[cc].push_back(p);
  for (int i = 1; i <= n; i++) {
    if (mf.f[p][i] != 0 && !vs[i]) {
      dfs(i);
      break;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    if (a[i] % 2) {
      mf.add(i, mf.t, 2, 1);
    } else {
      mf.add(mf.s, i, 2, 1);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (isPrime(a[i] + a[j])) {
        if (a[j] % 2) {
          mf.add(i, j, 1, 1);
        } else {
          mf.add(j, i, 1, 1);
        }
      }
    }
  }
  mf.calc();
  if (mf.x[mf.t] == n) {
    for (int i = 1; i <= n; i++) {
      if (!vs[i]) {
        cc++;
        dfs(i);
      }
    }
    printf("%d\n", cc);
    for (int i = 1; i <= cc; i++) {
      printf("%d ", ans[i].size());
      for (int p : ans[i]) {
        printf("%d ", p);
      }
      printf("\n");
    }
  } else {
    printf("Impossible");
  }
  return 0;
}