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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
int const N = 200 + 5, M = N * N, MAX = 1e4 + 20, inf = 1e9 + 20;
int n, a[N], second = 0, T = 1;
int ec, from[M], to[M], cap[M], nxt[M], head[N];
int F;
inline void add_edge(int u, int v, int c) {
  from[ec] = u, to[ec] = v, cap[ec] = c, nxt[ec] = head[u], head[u] = ec, ec++;
  from[ec] = v, to[ec] = u, cap[ec] = 0, nxt[ec] = head[v], head[v] = ec, ec++;
}
bool np[2 * MAX];
void pre() {
  np[0] = np[1] = 1;
  for (int i = 2; i < 2 * MAX; i++)
    for (int j = 2 * i; j < 2 * MAX; j += i) np[j] = 1;
}
int dist[N];
inline bool bfs(int second, int T) {
  memset(dist, 31, sizeof dist);
  dist[second] = 0;
  queue<int> que;
  que.push(second);
  while (que.size()) {
    int v = que.front();
    que.pop();
    for (int i = head[v]; i != -1; i = nxt[i]) {
      int u = to[i];
      if (cap[i] > 0 && dist[v] + 1 < dist[u]) {
        dist[u] = dist[v] + 1;
        que.push(u);
      }
    }
  }
  return dist[T] < n;
}
int par[N], ptr[N];
bool vis[N];
bool dfs(int v, int T) {
  vis[v] = 1;
  if (v == T) return 1;
  for (; ptr[v] != -1; ptr[v] = nxt[ptr[v]]) {
    int i = ptr[v];
    int u = to[i];
    if (cap[i] > 0 && dist[v] + 1 == dist[u] && !vis[u] && dfs(u, T)) {
      par[u] = i;
      return 1;
    }
  }
  return 0;
}
vector<int> g[N], vec[N];
void find_cyc(int v, int C) {
  vis[v] = 1;
  vec[C].push_back(v);
  for (int u : g[v])
    if (!vis[u]) find_cyc(u, C);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  memset(head, -1, sizeof head);
  pre();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] & 1)
      add_edge(i + 2, T, 2);
    else
      add_edge(second, i + 2, 2);
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (!(a[i] & 1) && !np[a[i] + a[j]]) add_edge(i + 2, j + 2, 1);
  while (bfs(second, T)) {
    for (int i = 0; i < n + 2; i++) ptr[i] = head[i];
    memset(vis, 0, sizeof vis);
    if (dfs(second, T)) {
      int mn = inf;
      for (int i = T; i != second; i = from[par[i]]) mn = min(mn, cap[par[i]]);
      F += mn;
      for (int i = T; i != second; i = from[par[i]]) {
        cap[par[i]] -= mn;
        cap[par[i] ^ 1] += mn;
      }
      memset(vis, 0, sizeof vis);
    }
  }
  if (F < n) return cout << "Impossible\n", 0;
  for (int i = 0; i < ec; i++) {
    int u = from[i], v = to[i];
    if (u > 1 && v > 1 && !(a[u - 2] & 1) && (a[v - 2] & 1) && !cap[i]) {
      g[u - 2].push_back(v - 2);
      g[v - 2].push_back(u - 2);
    }
  }
  int C = 0;
  memset(vis, 0, sizeof vis);
  for (int i = 0; i < n; i++)
    if (!vis[i]) find_cyc(i, C++);
  cout << C << '\n';
  for (int i = 0; i < C; i++) {
    cout << (int)vec[i].size() << ' ';
    for (int x : vec[i]) cout << x + 1 << ' ';
    cout << '\n';
  }
}