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 N = 205;
int read() {
  int x = 0, f = 1;
  char ch;
  while (!isdigit(ch = getchar())) (ch == '-') && (f = -f);
  for (x = ch ^ 48; isdigit(ch = getchar());
       x = (x << 3) + (x << 1) + (ch ^ 48))
    ;
  return x * f;
}
vector<int> vec[N];
int n, s, t, tim, cnt = 1, a[N], head[N], vis[N], level[N], is[20010];
struct edge {
  int v, w, nxt;
} e[N * N];
void add(int x, int y, int w) {
  e[++cnt] = (edge){y, w, head[x]};
  head[x] = cnt;
  e[++cnt] = (edge){x, 0, head[y]};
  head[y] = cnt;
}
void init(int n) {
  is[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (is[i]) continue;
    for (int j = i + i; j <= n; j += i) is[j] = 1;
  }
}
int bfs() {
  queue<int> Q;
  Q.push(s);
  memset(level, 0, sizeof(level));
  level[s] = 1;
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = head[x]; i; i = e[i].nxt) {
      int v = e[i].v, w = e[i].w;
      if (w && !level[v]) level[v] = level[x] + 1, Q.push(v);
    }
  }
  return level[t] != 0;
}
int dfs(int x, int flow) {
  if (x == t || !flow) return flow;
  int up = 0;
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].v, w = e[i].w;
    if (w && level[v] == level[x] + 1) {
      int tmp = dfs(v, min(flow, w));
      e[i].w -= tmp, e[i ^ 1].w += tmp;
      flow -= tmp, up += tmp;
      if (!flow) break;
    }
  }
  if (!up) level[x] = 0;
  return up;
}
int Dinic() {
  int maxflow = 0;
  while (bfs()) maxflow += dfs(s, 2e9);
  return maxflow;
}
void find(int x) {
  vis[x] = 1;
  vec[tim].push_back(x);
  for (int i = head[x]; i; i = e[i].nxt)
    if (!vis[e[i].v] && e[i | 1].w == 1) find(e[i].v);
}
int main() {
  n = read();
  s = 0;
  t = n + 1;
  init(20000);
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i <= n; ++i)
    if (a[i] & 1)
      add(i, t, 2);
    else
      add(s, i, 2);
  for (int i = 1; i <= n; ++i) {
    if (a[i] & 1) continue;
    for (int j = 1; j <= n; ++j)
      if (!is[a[i] + a[j]]) add(i, j, 1);
  }
  if (Dinic() != n) return puts("Impossible"), 0;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) ++tim, find(i);
  printf("%d\n", tim);
  for (int i = 1; i <= tim; ++i) {
    printf("%d ", vec[i].size());
    for (int j = 0; j < vec[i].size(); ++j) printf("%d ", vec[i][j]);
    puts("");
  }
  return 0;
}
