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;
struct Dinic {
  static const int N = 205, M = 50005;
  int n, s, t;
  struct edge {
    int v, w, f, nxt;
  } e[M];
  int head[N], cur[N], cnt;
  void addedge(int u, int v, int w, int f) {
    cnt++;
    e[cnt].v = v, e[cnt].w = w, e[cnt].f = f;
    e[cnt].nxt = head[u];
    head[u] = cnt;
  }
  void add(int u, int v, int w) {
    addedge(u, v, w, w);
    addedge(v, u, w, 0);
  }
  int dep[N], q[N];
  int bfs() {
    for (int i = 1; i <= n; i++) dep[i] = -1, cur[i] = head[i];
    dep[s] = 0;
    int l = 0, r = 1;
    q[0] = s;
    while (l < r) {
      int u = q[l++];
      for (int i = head[u]; i; i = e[i].nxt) {
        if (dep[e[i].v] == -1 && e[i].f > 0) {
          dep[e[i].v] = dep[u] + 1;
          q[r++] = e[i].v;
        }
      }
    }
    return dep[t] != -1;
  }
  int dfs(int u, int lim) {
    if (u == t || lim == 0) return lim;
    int flow = 0, f;
    for (int i = cur[u]; i; i = e[i].nxt) {
      cur[u] = i;
      if (dep[e[i].v] == dep[u] + 1 && (f = dfs(e[i].v, min(lim, e[i].f)))) {
        flow += f;
        lim -= f;
        e[i].f -= f;
        e[(i - 1 ^ 1) + 1].f += f;
        if (!lim) break;
      }
    }
    return flow;
  }
  int max_flow() {
    int mf = 0;
    while (bfs()) mf += dfs(s, 1e9);
    return mf;
  }
} g;
const int N = 205, M = 20005;
int a[N], n, p[M], vis[N];
vector<int> e[N], ans;
vector<vector<int>> k;
void dfs(int u) {
  ans.push_back(u);
  vis[u] = 1;
  for (auto v : e[u])
    if (!vis[v]) dfs(v);
}
int main() {
  for (int i = 2; i <= 20000; i++)
    if (!p[i])
      for (int j = i + i; j <= 20000; j += i) p[j] = 1;
  scanf("%d", &n);
  g.n = n + 2, g.s = n + 1, g.t = n + 2;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1) {
      g.add(i, n + 2, 2);
      continue;
    }
    g.add(n + 1, i, 2);
    for (int j = 1; j <= n; j++)
      if (a[j] & 1)
        if (!p[a[i] + a[j]]) g.add(i, j, 1);
  }
  if (g.max_flow() != n) {
    puts("Impossible");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1) continue;
    for (int j = g.head[i]; j; j = g.e[j].nxt) {
      if (g.e[j].v <= n && g.e[j].w != g.e[j].f) {
        e[i].push_back(g.e[j].v);
        e[g.e[j].v].push_back(i);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      ans.clear();
      dfs(i);
      k.push_back(ans);
    }
  }
  printf("%d\n", k.size());
  for (auto i : k) {
    printf("%d ", i.size());
    for (auto j : i) printf("%d ", j);
    puts("");
  }
  return 0;
}