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 Edge {
  int u, v;
  int cap, res;
  int next;
};
class Dinic {
 public:
  static const int N = 400;
  static const int M = 100000;
  static const int Inf = 0x3f3f3f3f;
  int n, m;
  Edge E[M + 10];
  int s[N + 10];
  int src, snk;
  int que[N + 10];
  bool vis[N + 10];
  int dist[N + 10];
  int cur[N + 10];
  void init() {
    memset(s, 0, sizeof(s));
    n = 0;
    m = 1;
  }
  void add_edge(int u, int v, int cap) {
    int p = ++m;
    E[p].u = u;
    E[p].v = v;
    E[p].res = E[p].cap = cap;
    E[p].next = s[u];
    s[u] = p;
  }
  void add_edge2(int u, int v, int cap) {
    add_edge(u, v, cap);
    add_edge(v, u, 0);
  }
  bool bfs() {
    memset(que, 0, sizeof(que));
    memset(vis, 0, sizeof(vis));
    memset(dist, 0, sizeof(dist));
    int f = 1, p = 2;
    que[1] = src;
    vis[src] = true;
    while (f < p) {
      int u = que[f++];
      for (int i = s[u]; i != 0; i = E[i].next)
        if (E[i].res > 0 && !vis[E[i].v]) {
          int v = E[i].v;
          que[p++] = v;
          vis[v] = true;
          dist[v] = dist[u] + 1;
        }
    }
    return vis[snk];
  }
  int dfs(int u, int neck) {
    if (u == snk || neck == 0) return neck;
    int sol = 0;
    for (int& i = cur[u]; i != 0; i = E[i].next)
      if (E[i].res > 0 && dist[E[i].v] == dist[u] + 1) {
        int tmp = dfs(E[i].v, min(neck, E[i].res));
        E[i].res -= tmp;
        E[i ^ 1].res += tmp;
        neck -= tmp;
        sol += tmp;
        if (neck == 0) break;
      }
    return sol;
  }
  int max_flow() {
    int sol = 0;
    while (bfs()) {
      memcpy(cur, s, sizeof(s));
      sol += dfs(src, Inf);
    }
    return sol;
  }
};
const int N = 200;
int n;
int a[N + 10];
Dinic G;
bool vis[N + 10];
vector<int> ans[N + 10];
int cnt;
bool judge_prime(int x) {
  int bound = (int)sqrt(x);
  for (int i = 2; i <= bound; i++)
    if (x % i == 0) return false;
  return true;
}
void dfs(int u, int k) {
  vis[u] = true;
  ans[k].push_back(u);
  for (int i = G.s[u]; i != 0; i = G.E[i].next)
    if (G.E[i].cap != G.E[i].res && !vis[G.E[i].v]) dfs(G.E[i].v, k);
}
void solve() {
  int s1 = 0, s2 = 0;
  for (int i = 1; i <= n; i++)
    if (a[i] % 2 == 1)
      s1++;
    else
      s2++;
  if (s1 != s2) {
    printf("Impossible\n");
    return;
  }
  G.init();
  G.n = n + 2;
  G.src = n + 1;
  G.snk = n + 2;
  for (int i = 1; i <= n; i++)
    if (a[i] % 2 == 1)
      G.add_edge2(G.src, i, 2);
    else
      G.add_edge2(i, G.snk, 2);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i] % 2 == 1 && a[j] % 2 == 0 && judge_prime(a[i] + a[j]))
        G.add_edge2(i, j, 1);
  int flow = G.max_flow();
  if (flow != n) {
    printf("Impossible\n");
    return;
  }
  memset(vis, 0, sizeof(vis));
  vis[G.src] = true;
  vis[G.snk] = true;
  cnt = 0;
  for (int i = G.s[G.src]; i != 0; i = G.E[i].next) {
    int v = G.E[i].v;
    if (!vis[v]) dfs(v, ++cnt);
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d ", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf("%d ", ans[i][j]);
    printf("\n");
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  solve();
  return 0;
}