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;
const int M = 1e5 + 10;
const int inf = 0x3f3f3f3f;
int n, cnt, E = 1, S = 201, T = 202, a[N], pri[M];
int fir[N], vis[N], dis[N], nex[M], arr[M], cap[M], use[N], can[N][N];
queue<int> Q;
vector<int> vec[N];
void Init() {
  for (int i = 2; i < M; i++) pri[i] = 1;
  for (int i = 2; i < M; i++)
    if (pri[i])
      for (int j = i + i; j < M; j += i) pri[j] = 0;
}
inline void Add_Edge(int x, int y, int c) {
  nex[++E] = fir[x];
  fir[x] = E;
  arr[E] = y;
  cap[E] = c;
  nex[++E] = fir[y];
  fir[y] = E;
  arr[E] = x;
  cap[E] = 0;
}
int Bfs() {
  memset(vis, 0, sizeof(vis));
  memset(dis, 0x3f, sizeof(dis));
  dis[S] = 0;
  Q.push(S);
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int i = fir[u]; i; i = nex[i])
      if (cap[i] && dis[arr[i]] == inf) {
        dis[arr[i]] = dis[u] + 1;
        Q.push(arr[i]);
      }
  }
  return dis[T] < inf;
}
int Dfs(int x, int flow) {
  if (x == T || !flow) return flow;
  int ans = 0;
  for (int i = fir[x]; i; i = nex[i])
    if (cap[i] && dis[arr[i]] == dis[x] + 1) {
      int del = Dfs(arr[i], min(flow, cap[i]));
      ans += del;
      cap[i] -= del;
      cap[i ^ 1] += del;
      flow -= del;
    }
  return ans;
}
void Dinic() {
  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++)
    if (a[i] & 1) {
      for (int j = 1; j <= n; j++)
        if (!(a[j] & 1) && pri[a[i] + a[j]]) {
          Add_Edge(i, j, 1);
        }
    }
  int ans = 0;
  while (Bfs()) ans += Dfs(S, inf);
  if (ans < n) puts("Impossible"), exit(0);
}
void Getans(int x) {
  for (int i = 1; i <= n; i++)
    if (can[x][i] && !use[i]) {
      use[i] = cnt;
      vec[cnt].push_back(i);
      Getans(i);
    }
}
int main() {
  Init();
  scanf("%d", &n);
  if (n & 1) return 0 * puts("Impossible");
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1)
      cnt++;
    else
      cnt--;
  }
  if (cnt) return 0 * puts("Impossible");
  Dinic();
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1) {
      for (int j = fir[i]; j; j = nex[j])
        if (arr[j] <= n && !cap[j]) can[i][arr[j]] = can[arr[j]][i] = 1;
    }
  }
  for (int i = 1; i <= n; i++)
    if (!use[i]) {
      use[i] = ++cnt;
      vec[cnt].push_back(i);
      Getans(i);
    }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d ", vec[i].size());
    for (int j = 0; j < vec[i].size(); j++) printf("%d ", vec[i][j]);
    putchar('\n');
  }
  return 0;
}
