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;
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[220];
vector<int> g1[220];
int level[220], iter[220];
bool vis[220];
vector<int> ret[220];
int a[220];
int n, s, t;
void add_edge(int from, int to, int cap) {
  g[from].push_back(edge(to, cap, g[to].size()));
  g[to].push_back(edge(from, 0, g[from].size() - 1));
}
bool bfs() {
  queue<int> q;
  memset(level, -1, sizeof(level));
  q.push(s);
  level[s] = 0;
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < g[v].size(); i++) {
      edge &e = g[v][i];
      if (e.cap > 0 && level[e.to] == -1) {
        level[e.to] = level[v] + 1;
        q.push(e.to);
      }
    }
  }
  return level[t] != -1;
}
int dfs(int v, int f) {
  if (v == t) return f;
  for (int &i = iter[v]; i < g[v].size(); i++) {
    edge &e = g[v][i];
    if (e.cap > 0 && level[e.to] == level[v] + 1) {
      int d = dfs(e.to, min(f, e.cap));
      if (d > 0) {
        e.cap -= d;
        g[e.to][e.rev].cap += d;
        return d;
      }
    }
  }
  return 0;
}
int max_flow() {
  int ans = 0, f;
  while (bfs()) {
    memset(iter, 0, sizeof(iter));
    while (true) {
      f = dfs(s, 1000000000);
      if (f == 0) break;
      ans += f;
    }
  }
  return ans;
}
void dfs1(int v, int cnt) {
  if (vis[v]) return;
  vis[v] = 1;
  ret[cnt].push_back(v);
  for (int i = 0; i < g1[v].size(); i++) dfs1(g1[v][i], cnt);
}
bool isp(int x) {
  if (x < 2) return false;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
int main() {
  scanf("%d", &n);
  if (n & 1) {
    puts("Impossible");
    return 0;
  }
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  s = n + 1;
  t = n + 2;
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1)
      add_edge(s, i, 2);
    else
      add_edge(i, t, 2);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i] % 2 == 1 && a[j] % 2 == 0 && isp(a[i] + a[j])) {
        add_edge(i, j, 1);
      }
    }
  }
  int flow = max_flow();
  if (flow != n) {
    puts("Impossible");
    return 0;
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1) {
      for (int j = 0; j < g[i].size(); j++) {
        edge &e = g[i][j];
        if (e.to > n) continue;
        if (e.cap == 0) {
          g1[i].push_back(e.to);
          g1[e.to].push_back(i);
        }
      }
    }
  }
  for (int i = 1; i <= n; i++)
    if (a[i] % 2 == 1 && !vis[i]) dfs1(i, ++cnt);
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d", ret[i].size());
    for (int j = 0; j < ret[i].size(); j++) printf(" %d", ret[i][j]);
    puts("");
  }
  return 0;
}
