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;
struct node {
  int from;
  int to;
  int w;
  int next;
} e[1515151];
vector<int> mp[2050];
int vis[2050];
int head[2050];
int cur[2050];
int divv[2050];
int lx[2050];
int ly[2050];
int a[205];
int Is_or[165550];
int n, cont, ss, tt, contx, conty;
void add(int from, int to, int w) {
  e[cont].from = from;
  e[cont].to = to;
  e[cont].w = w;
  e[cont].next = head[from];
  head[from] = cont++;
}
void init() {
  memset(Is_or, 0, sizeof(Is_or));
  for (int j = 2; j < sqrt(165540); j++) {
    if (Is_or[j] == 0)
      for (int k = j + j; k <= 165540; k += j) Is_or[k] = 1;
  }
}
void getmap() {
  cont = 0;
  ss = n + 1;
  tt = ss + 1;
  contx = 0, conty = 0;
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 == 0) {
      add(ss, i, 2);
      add(i, ss, 0);
      lx[contx++] = i;
    } else {
      add(i, tt, 2);
      add(tt, i, 0);
      ly[conty++] = i;
    }
  }
  for (int i = 0; i < contx; i++) {
    for (int j = 0; j < conty; j++) {
      if (Is_or[a[lx[i]] + a[ly[j]]] == 0) {
        add(lx[i], ly[j], 1);
        add(ly[j], lx[i], 0);
      }
    }
  }
}
int makedivv() {
  queue<int> s;
  s.push(ss);
  memset(divv, 0, sizeof(divv));
  divv[ss] = 1;
  while (!s.empty()) {
    int u = s.front();
    if (u == tt) return 1;
    s.pop();
    for (int i = head[u]; i != -1; i = e[i].next) {
      int v = e[i].to;
      int w = e[i].w;
      if (w && divv[v] == 0) {
        divv[v] = divv[u] + 1;
        s.push(v);
      }
    }
  }
  return 0;
}
int Dfs(int u, int maxflow, int tt) {
  if (u == tt) return maxflow;
  int ret = 0;
  for (int &i = cur[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    int w = e[i].w;
    if (w && divv[v] == divv[u] + 1) {
      int f = Dfs(v, min(maxflow - ret, w), tt);
      e[i].w -= f;
      e[i ^ 1].w += f;
      ret += f;
      if (ret == maxflow) return ret;
    }
  }
  return ret;
}
int Dinic() {
  int ans = 0;
  while (makedivv() == 1) {
    memcpy(cur, head, sizeof(head));
    ans += Dfs(ss, 0x3f3f3f3f, tt);
  }
  return ans;
}
void Dfss(int u, int tot) {
  mp[tot].push_back(u);
  vis[u] = 1;
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    if (vis[v] == 1) continue;
    int w = e[i].w;
    if (v != ss && v != tt) {
      if (i % 2 == 0 && w == 0) {
        Dfss(v, tot);
      }
      if (i % 2 == 1 && e[i ^ 1].w == 0) {
        Dfss(v, tot);
      }
    }
  }
}
int main() {
  init();
  while (~scanf("%d", &n)) {
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      mp[i].clear();
    }
    getmap();
    int tmp = Dinic();
    if (tmp == n) {
      int tot = 0;
      for (int i = 1; i <= n; i++) {
        if (vis[i] == 0) {
          Dfss(i, ++tot);
        }
      }
      printf("%d\n", tot);
      for (int i = 1; i <= tot; i++) {
        printf("%d", mp[i].size());
        for (int j = 0; j < mp[i].size(); j++) {
          printf(" %d", mp[i][j]);
        }
        printf("\n");
      }
    } else
      printf("Impossible\n");
  }
}