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 = 2e2 + 4;
const int MAXP = 5e4;
const int MAXV = MAXN + 4;
const int inf = 1e9;
bool isp[MAXP];
int a[MAXN];
int n;
int c[MAXV][MAXV];
int f[MAXV][MAXV];
int d[MAXV];
int q[MAXV];
int ptr[MAXV];
int qh, qt;
void bfs() {
  while (qh < qt) {
    int v = q[qh++];
    for (auto i = 0; i != n + 2; i++)
      if (c[v][i] - f[v][i] > 0 && d[i] == 0) {
        d[i] = d[v] + 1;
        q[qt++] = i;
      }
  }
}
int dfs(int v, int flow) {
  if (v == n + 1) return flow;
  if (flow == 0) return 0;
  for (int &u = ptr[v]; u < n + 2; u++) {
    if (d[u] == d[v] + 1 && c[v][u] - f[v][u] > 0) {
      int pushed = dfs(u, min(flow, c[v][u] - f[v][u]));
      if (pushed != 0) {
        f[v][u] += pushed;
        f[u][v] -= pushed;
        return pushed;
      }
    }
  }
  return 0;
}
int dinic() {
  int flow = 0;
  while (true) {
    memset(d, 0, sizeof(d));
    memset(ptr, 0, sizeof(ptr));
    qh = qt = 0;
    q[qt++] = n;
    d[n] = 1;
    bfs();
    if (d[n + 1] == 0) break;
    while (int pushed = dfs(n, inf)) flow += pushed;
  }
  return flow;
}
bool used[MAXN];
int lst[MAXN];
int sz = 0;
int ancnt = 0;
void mark(int v) {
  used[v] = true;
  lst[sz++] = v + 1;
  for (auto i = 0; i != n; i++) {
    int w = (a[i] & 1) ? f[i][v] : f[v][i];
    if (w > 0 && !used[i]) mark(i);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  memset(c, 0, sizeof(c));
  memset(f, 0, sizeof(f));
  memset(isp, 0, sizeof(isp));
  isp[2] = false;
  for (auto i = 2; i != MAXP; i++) {
    if (isp[i]) continue;
    for (long long j = i * 1ll * i; j < MAXP; j += i) isp[j] = true;
  }
  scanf("%d", &n);
  for (auto i = 0; i != n; i++) scanf("%d", &a[i]);
  for (auto i = 0; i != n; i++)
    for (auto j = i + 1; j != n; j++) {
      if (isp[a[i] + a[j]]) continue;
      if (a[i] & 1)
        c[i][j] = 1;
      else
        c[j][i] = 1;
    }
  for (auto i = 0; i != n; i++) {
    if (a[i] & 1)
      c[n][i] = 2;
    else
      c[i][n + 1] = 2;
  }
  int flow = dinic();
  if (flow != n) {
    printf("Impossible");
    return 0;
  }
  memset(used, 0, sizeof(used));
  for (auto i = 0; i != n; i++)
    if (!used[i]) {
      ancnt++;
      sz = 0;
      mark(i);
    }
  memset(used, 0, sizeof(used));
  printf("%d", ancnt);
  for (auto i = 0; i != n; i++)
    if (!used[i]) {
      printf("\n");
      sz = 0;
      mark(i);
      printf("%d ", sz);
      for (auto j = 0; j != sz; j++) printf("%d ", lst[j]);
    }
  return 0;
}
