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>
void fre() {
  freopen("c://test//input.in", "r", stdin);
  freopen("c://test//output.out", "w", stdout);
}
template <class T>
inline void gmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline void gmin(T &a, T b) {
  if (b < a) a = b;
}
using namespace std;
const int N = 220, M = 80000, Z = 1e9 + 7, maxint = 2147483647,
          ms31 = 522133279, ms63 = 1061109567, ms127 = 2139062143;
const double eps = 1e-8, PI = acos(-1.0);
bool e[20020];
void prime() {
  int top = 20000;
  for (int i = 2; i <= top; i++)
    if (e[i] == 0) {
      for (int j = i + i; j <= top; j += i) e[j] = 1;
    }
}
int n, id, ST, ED;
int a[N], first[N];
int w[M], c[M], cap[M], nxt[M];
void ins(int x, int y, int cap_) {
  id++;
  w[id] = y;
  cap[id] = cap_;
  nxt[id] = first[x];
  first[x] = id;
  id++;
  w[id] = x;
  cap[id] = 0;
  nxt[id] = first[y];
  first[y] = id;
}
int d[N];
bool bfs() {
  memset(d, -1, sizeof(d));
  d[ST] = 0;
  queue<int> q;
  q.push(ST);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int z = first[x]; z; z = nxt[z])
      if (cap[z]) {
        int y = w[z];
        if (d[y] == -1) {
          d[y] = d[x] + 1;
          if (y == ED) return 1;
          q.push(y);
        }
      }
  }
  return 0;
}
int dfs(int x, int all) {
  if (x == ED) return all;
  int use = 0;
  for (int z = first[x]; z; z = nxt[z])
    if (cap[z]) {
      int y = w[z];
      if (d[y] == d[x] + 1) {
        int tmp = dfs(y, min(cap[z], all - use));
        cap[z] -= tmp;
        cap[z ^ 1] += tmp;
        use += tmp;
        if (use == all) break;
      }
    }
  if (use == 0) d[x] = -1;
  return use;
}
int dinic() {
  int ans = 0;
  while (bfs()) ans += dfs(ST, n * 2);
  return ans;
}
bool vis[N];
int b[N][N], sum;
void find(int x) {
  vis[x] = 1;
  b[sum][++b[sum][0]] = x;
  if (a[x] & 1) {
    for (int z = first[x]; z; z = nxt[z]) {
      if (z % 2 == 0 && cap[z] == 0 && vis[w[z]] == 0) find(w[z]);
    }
  } else {
    for (int z = first[x]; z; z = nxt[z]) {
      if (z % 2 == 1 && cap[z] == 1 && vis[w[z]] == 0) find(w[z]);
    }
  }
}
void print() {
  memset(vis, 0, sizeof(vis));
  sum = 0;
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0) {
      sum++;
      b[sum][0] = 0;
      find(i);
    }
  printf("%d\n", sum);
  for (int i = 1; i <= sum; i++) {
    printf("%d ", b[i][0]);
    for (int j = 1; j <= b[i][0]; j++) printf("%d ", b[i][j]);
    puts("");
  }
}
int main() {
  prime();
  while (~scanf("%d", &n)) {
    ST = 0;
    ED = n + 1;
    memset(first, 0, sizeof(first));
    id = 1;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= n; i++) {
      if (a[i] & 1) {
        ins(ST, i, 2);
        for (int j = 1; j <= n; j++)
          if (e[a[i] + a[j]] == 0) ins(i, j, 1);
      } else
        ins(i, ED, 2);
    }
    if (n % 2 == 0 && dinic() == n)
      print();
    else
      printf("Impossible\n");
  }
  return 0;
}