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;
int s, flow;
vector<int> g[202];
int c[202][202], u[202][202], at[2001], l[2001];
bool bfs() {
  memset(l, -1, sizeof(l));
  l[0] = 0;
  queue<int> q;
  q.push(0);
  int x;
  while (!q.empty()) {
    x = q.front(), q.pop();
    for (int y : g[x])
      if ((c[x][y] > u[x][y]) && l[y] == -1) l[y] = l[x] + 1, q.push(y);
  }
  return l[s] + 1;
}
int dfs(int x, int ch) {
  if (x == s) return ch;
  for (; at[x] < g[x].size(); at[x]++) {
    int y = g[x][at[x]];
    if (l[y] != l[x] + 1 || c[x][y] == u[x][y]) continue;
    int f = dfs(y, min(ch, c[x][y] - u[x][y]));
    if (f) {
      u[x][y] += f, u[y][x] -= f;
      if (x == 0) flow += f;
      return f;
    }
  }
  return 0;
}
int a[201];
bitset<20001> p;
vector<vector<int>> v;
bool done[201];
int idk(int x) {
  int i;
  for (i = 1; i < s; i++)
    if (u[x][i] && !done[i]) return i;
  return 0;
}
int main() {
  int n, i, j;
  p.set();
  for (i = 2; i * i <= 20000; i++)
    if (p[i])
      for (j = i * i; j <= 20000; j += i) p[j] = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", a + i);
  s = n + 1;
  for (i = 1; i <= n; i++) {
    if (a[i] & 1)
      g[0].push_back(i), g[i].push_back(0), c[0][i] = 2;
    else
      g[i].push_back(s), g[s].push_back(i), c[i][s] = 2;
  }
  for (i = 1; i <= n; i++) {
    for (j = i + 1; j <= n; j++) {
      if (p[a[i] + a[j]]) {
        g[i].push_back(j), g[j].push_back(i);
        if (a[i] & 1)
          c[i][j] = 1;
        else
          c[j][i] = 1;
      }
    }
  }
  while (bfs()) {
    memset(at, 0, sizeof(at));
    while (dfs(0, 0x3f3f3f3f))
      ;
  }
  if (flow != n) {
    printf("Impossible");
    return 0;
  }
  for (i = 1; i <= n; i++) {
    if (done[i]) continue;
    int cur = i, lc = 0, t = 0;
    vector<int> th;
    while (cur) {
      th.push_back(cur);
      done[cur] = 1;
      cur = idk(cur);
    }
    v.push_back(th);
  }
  printf("%d\n", v.size());
  for (auto idek : v) {
    printf("%d ", idek.size());
    for (auto thing : idek) printf("%d ", thing);
    printf("\n");
  }
  getchar();
  getchar();
}