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 INF = 0x3F3F3F3F;
const int N = 205;
int n, m, x, y, a[N], v[N];
vector<int> f[N];
struct edge {
  int x, y, f;
  edge(int u, int v, int f) : x(u), y(v), f(f) {}
};
struct Netflow {
  vector<int> a[N];
  vector<edge> E;
  int n, m, flow;
  Netflow(int n) : n(n), m(0), flow(0) {}
  void addedge(int x, int y, int f) {
    E.push_back(edge(x, y, f));
    E.push_back(edge(y, x, 0));
    a[x].push_back(m++);
    a[y].push_back(m++);
  }
  int Maxflow(int s, int t) {
    int ans = flow, x = s, f = INF, tag[N], d[N], p[N], h[N], v[N];
    memset(tag, 0, sizeof(tag));
    memset(d, 0, sizeof(d));
    memset(h, 0, sizeof(h));
    memset(v, 0, sizeof(v));
    v[0] = n;
    while (d[s] < n) {
      h[x] = f;
      bool flag = false;
      for (int i = tag[x]; i < a[x].size(); i++) {
        edge &e = E[a[x][i]];
        if ((e.f > 0) && (d[x] == d[e.y] + 1)) {
          flag = true;
          tag[x] = i;
          f = min(e.f, f);
          p[e.y] = a[x][i];
          x = e.y;
          if (x == t) {
            ans += f;
            while (x != s) {
              E[p[x]].f -= f;
              E[p[x] ^ 1].f += f;
              x = E[p[x]].x;
            }
            f = INF;
          }
          break;
        }
      }
      if (flag) continue;
      int t = n;
      for (int i = 0; i < a[x].size(); i++)
        if ((E[a[x][i]].f > 0) && (d[E[a[x][i]].y] < t)) {
          t = d[E[a[x][i]].y];
          tag[x] = i;
        }
      if (--v[d[x]] == 0) break;
      v[d[x] = t + 1]++;
      if (x != s) {
        x = E[p[x]].x;
        f = h[x];
      }
    }
    return flow = ans;
  }
};
bool prime(int p) {
  for (int i = 2; i * i <= p; i++)
    if (p % i == 0) return false;
  return true;
}
int main() {
  scanf("%d", &n);
  Netflow g(n + 2);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int i = 0; i < n; i++) {
    if (a[i] % 2 == 0) {
      g.addedge(0, i + 1, 2);
      for (int j = 0; j < n; j++)
        if (prime(a[i] + a[j])) g.addedge(i + 1, j + 1, 1);
    } else
      g.addedge(i + 1, n + 1, 2);
  }
  g.Maxflow(0, n + 1);
  for (int i = 0; i < g.a[0].size(); i++) {
    edge e = g.E[g.a[0][i]];
    int x = e.y;
    for (int j = 0; j < g.a[x].size(); j++) {
      edge t = g.E[g.a[x][j]];
      if ((t.f == 0) && (t.y != 0)) {
        f[x].push_back(t.y);
        f[t.y].push_back(t.x);
      }
    }
  }
  for (int i = 1; i <= n; i++)
    if (f[i].size() < 2) {
      puts("Impossible");
      return 0;
    }
  vector<vector<int> > ans;
  for (int i = 1; i <= n; i++) {
    if (v[i]) continue;
    vector<int> t;
    for (int x = i; !v[x]; x = ((v[f[x][0]]) ? f[x][1] : f[x][0])) {
      t.push_back(x);
      v[x] = 1;
    }
    ans.push_back(t);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) {
    printf("%d", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf(" %d", ans[i][j]);
    puts("");
  }
  return 0;
}