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;
const int N = 210;
int n, cnt = -1, ans = 0, flg = 0, res = 0;
int head[N], a[N], g[N], vis[N];
struct data {
  int to, nt, f, c;
} A[N * N];
vector<int> B[N], qx[N];
queue<int> q;
void add(int x, int y, int c) {
  A[++cnt].to = y, A[cnt].c = c, A[cnt].f = 0, A[cnt].nt = head[x],
  head[x] = cnt;
  A[++cnt].to = x, A[cnt].c = 0, A[cnt].f = 0, A[cnt].nt = head[y],
  head[y] = cnt;
}
bool check(int x) {
  int len = sqrt(x);
  for (int i = 2; i <= len; ++i)
    if (x % i == 0) return false;
  return true;
}
int mf[N], p[N];
bool bfs() {
  memset(mf, 0, sizeof mf), memset(p, 0, sizeof p);
  mf[0] = 10000, p[0] = -1;
  q.push(0);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; i != -1; i = A[i].nt) {
      int pos = A[i].c - A[i].f, to = A[i].to;
      if (!mf[to] && pos > 0) {
        mf[to] = min(pos, mf[u]);
        p[to] = i;
        q.push(to);
        if (to == n + 1) {
          while (!q.empty()) q.pop();
          break;
        }
      }
    }
  }
  int be = n + 1;
  if (!mf[n + 1]) return false;
  for (; be;) {
    A[p[be]].f += mf[n + 1];
    A[p[be] ^ 1].f -= mf[n + 1];
    be = A[p[be] ^ 1].to;
  }
  ans += mf[n + 1];
  return mf[n + 1];
}
void pd() {
  int res = 0;
  for (int i = head[0]; i != -1; i = A[i].nt)
    if (A[i].f == 2) res++;
  for (int i = head[n + 1]; i != -1; i = A[i].nt)
    if (A[i].f == -2) res++;
  if (res == n) return;
  printf("Impossible\n");
  exit(0);
}
void dfs(int u) {
  qx[res].push_back(u);
  vis[u] = 1;
  int len = B[u].size();
  for (int i = 0; i <= len - 1; ++i) {
    int to = B[u][i];
    if (!vis[to])
      g[to] = u, dfs(to);
    else if (to != g[u])
      flg = 1;
    else
      continue;
    return;
  }
}
int main() {
  memset(head, -1, sizeof head);
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    if (a[i] & 1) {
      add(0, i, 2);
      for (int j = 1; j <= n; ++j)
        if (check(a[i] + a[j])) add(i, j, 1);
    } else
      add(i, n + 1, 2);
  }
  bfs();
  for (; bfs();)
    ;
  pd();
  for (int u = 1; u <= n; ++u)
    for (int i = head[u]; i; i = A[i].nt) {
      if (A[i].c == A[i].f && A[i].f == 1) {
        B[u].push_back(A[i].to);
        B[A[i].to].push_back(u);
      }
    }
  for (int i = 1; i <= n; ++i) {
    if (a[i] % 2 && !vis[i]) g[i] = i, ++res, flg = 0, dfs(i);
  }
  printf("%d\n", res);
  for (int i = 1; i <= res; ++i) {
    int len = qx[i].size();
    printf("%d ", len);
    for (int j = 0; j <= len - 1; ++j) printf("%d ", qx[i][j]);
    puts("");
  }
  return 0;
}