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 MAXN = 200 + 10;
const int MAXM = MAXN * MAXN * 2;
const int INF = 0x7FFFFFFF >> 2;
struct Edge {
  int next, to, w;
} g[MAXM];
int gn = 1, start[MAXN];
inline void AddEdge(int a, int b, int c) {
  gn++;
  g[gn].next = start[a];
  start[a] = gn;
  g[gn].to = b;
  g[gn].w = c;
  gn++;
  g[gn].next = start[b];
  start[b] = gn;
  g[gn].to = a;
  g[gn].w = 0;
}
int a[MAXN], S, T, n, ans, cnt, h[MAXN], vh[MAXN], N, now[MAXN],
    ansc[MAXN][MAXN];
bool vis[MAXN];
bool isPrime(int x) {
  if (x % 2 == 0) return 0;
  for (int i = 3; i * i <= x; i += 2)
    if (x % i == 0) return 0;
  return 1;
}
int Aug(int p, int flow) {
  if (p == T) return flow;
  int ret = 0;
  for (int i = now[p]; i; i = g[i].next)
    if (g[i].w > 0) {
      int v = g[i].to;
      if (h[p] == h[v] + 1) {
        int now = Aug(v, min(flow - ret, g[i].w));
        ret += now;
        g[i].w -= now;
        g[i ^ 1].w += now;
        if (ret == flow) return ret;
      }
    }
  if (!--vh[h[p]]) h[S] = N;
  vh[++h[p]]++;
  now[p] = start[p];
  return ret;
}
void dfs(int p) {
  ansc[cnt][++ansc[cnt][0]] = p;
  vis[p] = 1;
  if (a[p] & 1) {
    for (int i = start[p]; i; i = g[i].next)
      if (g[i ^ 1].w > 0 && !vis[g[i].to]) dfs(g[i].to);
  } else
    for (int i = start[p]; i; i = g[i].next)
      if (g[i].w > 0 && !vis[g[i].to]) dfs(g[i].to);
}
int main() {
  scanf("%d", &n);
  S = n + 1;
  T = S + 1;
  N = T;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i)
    if (a[i] % 2 == 0) cnt++;
  if (cnt != n - cnt) {
    puts("Impossible");
    return 0;
  }
  for (int i = 1; i <= n; ++i)
    if (a[i] & 1) {
      AddEdge(S, i, 2);
      for (int j = 1; j <= n; ++j)
        if (isPrime(a[i] + a[j])) AddEdge(i, j, 1);
    } else
      AddEdge(i, T, 2);
  vh[0] = N;
  for (int i = 1; i <= N; ++i) now[i] = start[i];
  while (h[S] < N) ans += Aug(S, INF);
  if (ans != n) {
    puts("Impossible");
    return 0;
  }
  cnt = 0;
  vis[S] = 1;
  vis[T] = 1;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) {
      ++cnt;
      dfs(i);
    }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; ++i) {
    printf("%d ", ansc[i][0]);
    for (int j = 1; j < ansc[i][0]; ++j) printf("%d ", ansc[i][j]);
    printf("%d\n", ansc[i][ansc[i][0]]);
  }
  return 0;
}