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 = 310;
const int M = 210005;
int num[N], r[N], sz = 0, sr = 0, ord1[N], ord2[N];
bool divide[40005], vis[N];
int head[N], Next[M], V[M], C[M], cnt = 1, dep[N];
vector<int> ans[N / 3];
inline void prime() {
  for (int i = 2; i <= 40000; i++) {
    if (!divide[i]) {
      for (int j = 2; j <= 40000 / i; j++) divide[i * j] = 1;
    }
  }
}
inline void add(int x, int y, int w) {
  V[++cnt] = y;
  C[cnt] = w;
  Next[cnt] = head[x];
  head[x] = cnt;
}
inline void link(int x, int y, int w) {
  add(x, y, w);
  add(y, x, 0);
}
queue<int> Q;
inline bool BFS(int s, int t) {
  memset(dep, -1, sizeof(dep));
  dep[s] = 0;
  Q.push(s);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = head[x]; i; i = Next[i]) {
      if (dep[V[i]] == -1 && C[i]) {
        dep[V[i]] = dep[x] + 1;
        Q.push(V[i]);
      }
    }
  }
  return (dep[t] != -1);
}
inline int DFS(int x, int t, int flow) {
  if (!flow || x == t) return flow;
  int res = 0, d;
  for (int i = head[x]; i; i = Next[i]) {
    if (dep[V[i]] == dep[x] + 1 && C[i]) {
      d = DFS(V[i], t, min(flow, C[i]));
      C[i] -= d;
      C[i ^ 1] += d;
      res += d;
      flow -= d;
      if (!flow) break;
    }
  }
  if (!res) dep[x] = -1;
  return res;
}
inline void write(int x, int st, int la, int id) {
  ans[id].push_back(ord1[x]);
  vis[x] = 1;
  for (int i = head[x]; i; i = Next[i]) {
    if (V[i] == 0 || V[i] == sz * 2 + 1 || V[i] == st || V[i] == la ||
        vis[V[i]])
      continue;
    if (x <= sz && C[i]) continue;
    if (x > sz && !C[i]) continue;
    write(V[i], st, x, id);
    break;
  }
}
int main() {
  int n, k;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &k);
    if (k % 2 == 1)
      num[++sz] = k, ord1[sz] = i;
    else
      r[++sr] = k, ord2[sr] = i;
  }
  if (sz != sr) {
    printf("Impossible\n");
    return 0;
  }
  prime();
  for (int i = sz + 1; i <= n; i++) num[i] = r[i - sz], ord1[i] = ord2[i - sz];
  for (int i = 1; i <= sz; i++) link(0, i, 2);
  for (int i = sz + 1; i <= n; i++) link(i, n + 1, 2);
  for (int i = 1; i <= sz; i++) {
    for (int j = sz + 1; j <= n; j++) {
      if (!divide[num[i] + num[j]]) link(i, j, 1);
    }
  }
  int check = 0;
  while (BFS(0, n + 1)) check += DFS(0, n + 1, 2e9);
  if (check != n) {
    printf("Impossible\n");
    return 0;
  }
  int cnt = 0;
  for (int i = 1; i <= sz; i++) {
    if (!vis[i]) {
      cnt++;
      write(i, i, 0, cnt);
    }
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    int s = ans[i].size();
    printf("%d ", s);
    for (int j = 0; j < s; j++) {
      printf("%d", ans[i][j]);
      if (j != s - 1)
        printf(" ");
      else
        printf("\n");
    }
  }
  return 0;
}
