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.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 210;
struct Edge {
  int to, cap, rev;
  Edge(int _to = 0, int _cap = 0, int _rev = 0)
      : to(_to), cap(_cap), rev(_rev) {}
};
vector<Edge> g[maxn];
int level[maxn], iter[maxn];
inline void add_edge(int u, int v, int cap) {
  g[u].push_back(Edge(v, cap, (int)g[v].size()));
  g[v].push_back(Edge(u, 0, (int)g[u].size() - 1));
}
bool bfs(int source, int sink) {
  memset(level, -1, sizeof level);
  memset(iter, 0, sizeof iter);
  queue<int> q;
  q.push(source);
  level[source] = 0;
  while (!q.empty()) {
    int from = q.front();
    q.pop();
    for (int i = 0; i < (int)g[from].size(); i++) {
      Edge& e = g[from][i];
      if (e.cap > 0 && level[e.to] == -1) {
        level[e.to] = level[from] + 1;
        q.push(e.to);
      }
    }
  }
  return level[sink] != -1;
}
int find_flow(int from, int sink, int max_flow) {
  if (from == sink) return max_flow;
  for (int& i = iter[from]; i < (int)g[from].size(); i++) {
    Edge& e = g[from][i];
    if (e.cap > 0 && level[e.to] > level[from]) {
      int f = find_flow(e.to, sink, min(max_flow, e.cap));
      if (f > 0) {
        e.cap -= f;
        g[e.to][e.rev].cap += f;
        return f;
      }
    }
  }
  return 0;
}
inline int find_max_flow(int source, int sink) {
  int max_flow = 0, flow;
  while (bfs(source, sink)) {
    while ((flow = find_flow(source, sink, 0x3f3f3f3f)) > 0) max_flow += flow;
  }
  return max_flow;
}
bool notPrime[20010];
vector<int> prime;
void sieve() {
  notPrime[0] = notPrime[1] = true;
  for (int i = 2; i <= 20000; i++) {
    if (!notPrime[i]) prime.push_back(i);
    for (int j = 0; j < (int)prime.size() && (long long)prime[j] * i <= 20000;
         j++) {
      int d = prime[j] * i;
      notPrime[d] = true;
      if (i % prime[j] == 0) break;
    }
  }
}
int a[210], pre[210], nxt[210];
bool vis[210];
vector<int> ans[210];
int ans_size;
int n;
void get_ans(int from) {
  ans_size++;
  while (!vis[from]) {
    ans[ans_size].push_back(from);
    vis[from] = true;
    for (int i = 0; i < (int)g[from].size(); i++) {
      Edge& e = g[from][i];
      if (1 <= e.to && e.to <= n && (a[from] & 1) != (a[e.to] & 1) &&
          !vis[e.to]) {
        if ((a[from] & 1) && e.cap == 1) {
          from = e.to;
          break;
        } else if ((a[from] % 2 == 0) && e.cap == 0) {
          from = e.to;
          break;
        }
      }
    }
  }
}
int main() {
  sieve();
  scanf("%d", &n);
  int source = 0, sink = n + 1;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1)
      add_edge(i, sink, 2);
    else {
      add_edge(source, i, 2);
      for (int j = 1; j <= n; j++)
        if (a[j] & 1 && !notPrime[a[j] + a[i]]) add_edge(i, j, 1);
    }
  }
  if (find_max_flow(source, sink) != n) puts("Impossible"), exit(0);
  for (int i = 1; i <= n; i++)
    if (!vis[i]) get_ans(i);
  printf("%d\n", ans_size);
  for (int i = 1; i <= ans_size; i++) {
    printf("%d ", (int)ans[i].size());
    for (int j = 0; j < (int)ans[i].size(); j++) printf("%d ", ans[i][j]);
    puts("");
  }
  return 0;
}
