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;
const int Inf = 1e9;
const int Maxn = 234;
bool isp[22233];
void getp() {
  isp[1] = 1;
  for (int i = 2; i <= 20000; i++)
    if (!isp[i]) {
      for (int j = i + i; j <= 20000; j += i) isp[j] = 1;
    }
}
int n, s, t, ne, ans;
int in[Maxn], h[Maxn], cur[Maxn];
vector<int> G[Maxn];
pair<int, int> e[200020];
bool done[Maxn];
bool bfs() {
  for (int i = s; i <= t; i++) h[i] = i == s ? 0 : Inf;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < G[u].size(); i++) {
      if (!e[G[u][i]].first) continue;
      int v = e[G[u][i]].second;
      if (h[v] == Inf) h[v] = h[u] + 1, q.push(v);
    }
  }
  return h[t] != Inf;
}
int dfs(int u, int a) {
  if (u == t || !a) return a;
  int flow = 0, b;
  for (int &i = cur[u]; i < G[u].size() && a; i++) {
    int id = G[u][i], v = e[id].second;
    if (h[v] == h[u] + 1 && (b = dfs(v, min(a, e[id].first)))) {
      e[id ^ 1].first += b;
      e[id].first -= b;
      a -= b;
      flow += b;
    }
  }
  return flow;
}
void add(int u, int v, int c) {
  e[ne] = pair<int, int>(c, v);
  G[u].push_back(ne++);
  e[ne] = pair<int, int>(0, u);
  G[v].push_back(ne++);
}
vector<int> V[Maxn];
int cnt = 0;
void dfs(int u) {
  done[u] = 1;
  V[cnt].push_back(u);
  for (int i = 0; i < G[u].size(); i++) {
    int v = e[G[u][i]].second, c = e[G[u][i]].first;
    if (v != s && v != t && !done[v] && ((c == 1) ^ (in[u] & 1))) {
      dfs(v);
      return;
    }
  }
}
void pt() {
  for (int i = 1; i <= n; i++)
    if (!done[i]) {
      V[++cnt].clear();
      dfs(i);
    }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d", (int)V[i].size());
    for (int j = 0; j < V[i].size(); j++) printf(" %d", V[i][j]);
    puts("");
  }
}
int main() {
  getp();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", in + i);
  s = 0;
  t = n + 1;
  for (int i = 1; i <= n; i++)
    if (in[i] & 1) {
      add(s, i, 2);
      for (int j = 1; j <= n; j++)
        if (!isp[in[j] + in[i]]) add(i, j, 1);
    } else
      add(i, t, 2);
  while (bfs()) {
    memset(cur, 0, sizeof(cur));
    ans += dfs(s, Inf);
  }
  if (ans != n || !ans)
    puts("Impossible");
  else
    pt();
}