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;
int a[222];
int p[20010];
int q[222 * 222];
int qf, qb, fl[222];
int pre[222 * 222], dst[222 * 222];
int cap[222][222];
int adj[222][222];
int deg[222 * 222], cnt;
vector<int> A[222], An[222];
void dfs(int u, int fa = 0) {
  fl[u] = 1;
  An[cnt].push_back(u);
  for (int v : A[u])
    if (v != fa) {
      if (fl[v]) continue;
      dfs(v, u);
    }
}
void add(int u, int v, int c) {
  adj[u][deg[u]++] = v;
  adj[v][deg[v]++] = u;
  cap[u][v] = c;
  cap[v][u] = 0;
}
int max_flow(int src, int tar) {
  int ans = 0;
  while (1) {
    qb = qf = 0;
    q[qb++] = src;
    memset(pre, -1, sizeof pre);
    while (qb > qf && pre[tar] == -1) {
      int u = q[qf++];
      for (int i = 0; i < deg[u]; i++) {
        int v = adj[u][i];
        if (pre[v] != -1 || cap[u][v] == 0) continue;
        q[qb++] = v;
        pre[v] = u;
      }
    }
    if (pre[tar] == -1) return ans;
    int bot = 0x3f3f3f3f;
    for (int t = tar; t != src; t = pre[t]) {
      bot = min(bot, cap[pre[t]][t]);
    }
    ans += bot;
    for (int t = tar; t != src; t = pre[t]) {
      cap[pre[t]][t] -= bot;
      cap[t][pre[t]] += bot;
    }
  }
}
int main() {
  for (int i = 2; i <= 20000; i++) {
    if (p[i]) continue;
    for (int j = i * 2; j <= 20000; j += i) p[j] = 1;
  }
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  if (n & 1) {
    puts("Impossible");
    return 0;
  }
  int N = n + 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1)
      add(0, i, 2);
    else
      add(i, N, 2);
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 == 0) continue;
    for (int j = 1; j <= n; j++) {
      if (a[j] & 1) continue;
      if (!p[a[i] + a[j]]) add(i, j, 1);
    }
  }
  int mx = max_flow(0, N);
  if (mx != n) {
    puts("Impossible");
    return 0;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i] & 1 and a[j] % 2 == 0 and cap[j][i])
        A[i].push_back(j), A[j].push_back(i);
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (fl[i]) continue;
    cnt++;
    dfs(i);
  }
  cout << cnt << endl;
  for (int i = 1; i <= cnt; i++) {
    printf("%d", An[i].size());
    for (int v : An[i]) printf(" %d", v);
    puts("");
  }
  return 0;
}