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 N = 205, inf = 1e9;
int n, a[N];
int tot, lnk[N], fro[N * N * 2], nxt[N * N * 2], son[N * N * 2], c[N * N * 2],
    ori[N * N * 2];
int S, T, dep[N], cur[N];
queue<int> q;
int cnt;
bool vis[N];
vector<int> ans[N];
bool isprm(int x) {
  if (x == 2 || x == 3 || x == 5 || x == 7) return 1;
  for (int i = 2; i <= sqrt(x); ++i)
    if (x % i == 0) return 0;
  return 1;
}
void add(int x, int y, int z) {
  nxt[++tot] = lnk[x], lnk[x] = tot;
  fro[tot] = x, son[tot] = y, c[tot] = ori[tot] = z;
}
struct Dinic {
  bool bfs(int x = 0) {
    memset(dep, 0, sizeof dep), dep[S] = 1;
    for (; !q.empty(); q.pop())
      ;
    q.push(S);
    for (; !q.empty();) {
      x = q.front(), q.pop();
      for (int j = lnk[x]; j; j = nxt[j]) {
        if (!dep[son[j]] && c[j]) {
          dep[son[j]] = dep[x] + 1;
          q.push(son[j]);
        }
      }
    }
    return dep[T];
  }
  int dfs(int x, int f, int ret = 0) {
    if (x == T) return f;
    for (int &j = cur[x]; j; j = nxt[j]) {
      if (dep[son[j]] == dep[x] + 1 && c[j]) {
        int delta = dfs(son[j], min(f, c[j]));
        c[j] -= delta, c[j ^ 1] += delta;
        ret += delta, f -= delta;
      }
    }
    return ret;
  }
  int answer(int ret = 0) {
    for (; bfs();) {
      for (int i = S; i <= T; ++i) cur[i] = lnk[i];
      ret += dfs(S, inf);
    }
    return ret;
  }
} network;
void dfs(int x, int s) {
  vis[x] = 1, ans[cnt].push_back(x);
  if (a[x] & 1) {
    for (int j = lnk[x]; j; j = nxt[j]) {
      if (c[j] != 0 || son[j] == S || son[j] == T) continue;
      if (!vis[son[j]]) {
        dfs(son[j], s);
        break;
      }
    }
  } else {
    for (int j = lnk[x]; j; j = nxt[j]) {
      if (c[j ^ 1] != 0 || son[j] == S || son[j] == T) continue;
      if (!vis[son[j]]) {
        dfs(son[j], s);
        break;
      }
    }
  }
}
int main() {
  cin >> n, tot = 1, S = 0, T = n + 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] & 1)
      add(S, i, 2), add(i, S, 0);
    else
      add(i, T, 2), add(T, i, 0);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j)
      if (isprm(a[i] + a[j]) && a[i] % 2 == 1 && a[j] % 2 == 0)
        add(i, j, 1), add(j, i, 0);
  }
  int fl = network.answer();
  if (fl != n) return puts("Impossible"), 0;
  for (int i = 1; i <= n; ++i)
    if (a[i] & 1) {
      if (!vis[i]) ++cnt, dfs(i, i);
    }
  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]);
    puts("");
  }
  return 0;
}
