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>
inline int mymin(int x, int y) { return x < y ? x : y; }
bool isprime[22000];
inline void init_prime() {
  memset(isprime, true, sizeof(isprime));
  for (int i = 2; i <= (int)sqrt(20000 * 1.0); ++i)
    if (isprime[i])
      for (int j = 2; i * j <= 20000; ++j) isprime[i * j] = false;
}
struct edge {
  int to, c, next;
} e[22000];
int n, id[220], a[220], head[220], h[220], q[220], cnt, tot, num[55],
    path[55][220], cur[220];
bool f[220];
inline void print() { printf("Impossible\n"); }
inline void ins(int u, int v, int c) {
  e[++cnt].to = v;
  e[cnt].c = c;
  e[cnt].next = head[u];
  head[u] = cnt;
  e[++cnt].to = u;
  e[cnt].c = 0;
  e[cnt].next = head[v];
  head[v] = cnt;
}
int dfs(int u, int in) {
  if (u == n + 1 || !in) return in;
  int out = 0;
  for (int &i = cur[u]; i; i = e[i].next) {
    int v = e[i].to, c = e[i].c;
    if (h[v] == h[u] + 1 && c) {
      int flow = dfs(v, mymin(in - out, c));
      if (!flow)
        h[v] = 0;
      else
        out += flow;
      e[i].c -= flow;
      e[i ^ 1].c += flow;
      if (in == out) return out;
    }
  }
  return out;
}
inline bool bfs() {
  int op = 0, cl = 0;
  memset(h, 0, sizeof(h));
  h[q[++cl] = 0] = 1;
  while (op < cl) {
    int u = q[++op];
    if (u == n + 1) break;
    for (int i = head[u]; i; i = e[i].next) {
      int v = e[i].to, c = e[i].c;
      if (!h[v] && c) h[q[++cl] = v] = h[u] + 1;
    }
  }
  return h[n + 1];
}
inline int Dinic() {
  int rec = 0;
  while (bfs()) {
    memcpy(cur, head, sizeof(cur));
    rec += dfs(0, n);
  }
  return rec;
}
int dfs_path(int x) {
  path[tot][++num[tot]] = x;
  int y;
  for (int i = head[x]; i; i = e[i].next) {
    if (e[i].c) continue;
    y = e[i].to;
    if (f[y]) continue;
    f[y] = true;
    path[tot][++num[tot]] = y;
    break;
  }
  for (int i = head[y]; i; i = e[i].next) {
    if (e[i ^ 1].c) continue;
    x = e[i].to;
    if (f[x]) continue;
    f[x] = true;
    dfs_path(x);
  }
}
int main() {
  scanf("%d", &n);
  if (n & 1) {
    print();
    return 0;
  }
  int cnt1 = 0, cnt2 = n >> 1;
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    if (x & 1) {
      a[++cnt1] = x;
      id[cnt1] = i;
    } else {
      a[++cnt2] = x;
      id[cnt2] = i;
    }
    if (cnt1 > n >> 1 || cnt2 > n) {
      print();
      return 0;
    }
  }
  init_prime();
  memset(head, 0, sizeof(head));
  cnt = 1;
  for (int i = 1; i <= (n >> 1); ++i) {
    ins(0, id[i], 2);
    for (int j = (n >> 1) + 1; j <= n; ++j)
      if (isprime[a[i] + a[j]]) ins(id[i], id[j], 1);
  }
  for (int i = (n >> 1) + 1; i <= n; ++i) ins(id[i], n + 1, 2);
  int maxflow = Dinic();
  if (maxflow != n) {
    print();
    return 0;
  }
  memset(f, false, sizeof(f));
  f[0] = f[n + 1] = true;
  tot = 0;
  for (int i = 1; i <= (n >> 1); ++i)
    if (!f[id[i]]) {
      f[id[i]] = true;
      ++tot;
      num[tot] = 0;
      dfs_path(id[i]);
    }
  printf("%d\n", tot);
  for (int i = 1; i <= tot; ++i) {
    printf("%d", num[i]);
    for (int j = 1; j <= num[i]; ++j) printf(" %d", path[i][j]);
    if (i != tot) printf("\n");
  }
  return 0;
}