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 maxN = 402, maxE = 4e4, src = 400, snk = 401;
int head[maxN], headcpy[maxN], nxt[maxE], to[maxE], cap[maxE], n, arr[210],
    edgeCnt, vis[maxN], m;
vector<int> vec[210];
void addedge(int u, int v, int c) {
  nxt[edgeCnt] = head[u];
  head[u] = edgeCnt;
  cap[edgeCnt] = c;
  to[edgeCnt++] = v;
}
void addbiedge(int u, int v, int c) {
  addedge(u, v, c);
  addedge(v, u, 0);
}
int Rank[maxN];
int ddfs(int cur = src) {
  if (cur == snk) return 1;
  for (int &i = headcpy[cur]; i != -1; i = nxt[i]) {
    int t = to[i];
    if (!cap[i] || Rank[t] != Rank[cur] + 1) continue;
    if (ddfs(t)) {
      --cap[i];
      ++cap[i ^ 1];
      return 1;
    }
  }
  return 0;
}
int Q[maxN];
bool dbfs() {
  memset(Rank, -1, sizeof Rank);
  int Qi = 0;
  Q[Qi++] = src;
  Rank[src] = 0;
  for (int j = 0; j < Qi; ++j) {
    int cur = Q[j];
    int r = Rank[cur];
    for (int i = head[cur]; i != -1; i = nxt[i]) {
      int t = to[i];
      if (!cap[i] || Rank[t] != -1) continue;
      Rank[t] = r + 1;
      Q[Qi++] = t;
      if (t == snk) return 1;
    }
  }
  return 0;
}
void dfs(int cur) {
  if (vis[cur] || cur == src || cur == snk) return;
  vis[cur] = 1;
  vec[m].push_back(cur);
  for (int i = head[cur]; i != -1; i = nxt[i]) {
    if ((cur < 200) == (!cap[i])) dfs(to[i]);
  }
}
int dinic() {
  int res = 0;
  while (dbfs()) {
    for (int i = 0; i < maxN; ++i) headcpy[i] = head[i];
    while (ddfs()) ++res;
  }
  return res;
}
int main() {
  bitset<100001> ip;
  ip.set();
  ip[0] = ip[1] = 0;
  for (long long i = 2; i * i <= 100000; i++)
    if (ip[i])
      for (long long j = i * i; j <= 100000; j += i) ip[j] = 0;
  memset(head, -1, sizeof head);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &arr[i]);
    if (arr[i] & 1)
      addbiedge(200 + i, snk, 2);
    else
      addbiedge(src, i, 2);
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if ((arr[j] & 1) && ip[arr[i] + arr[j]]) addbiedge(i, 200 + j, 1);
  if (dinic() != n) return puts("Impossible");
  for (int i = 0; i < n; ++i)
    if (!(arr[i] & 1) && !vis[i]) dfs(i), m++;
  cout << m << endl;
  for (int i = 0; i < m; ++i) {
    printf("%d", vec[i].size());
    for (int j = 0; j < vec[i].size(); ++j)
      printf(" %d", vec[i][j] + 1 - (vec[i][j] > 199 ? 200 : 0));
    puts("");
  }
}
