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 n, m, x1, x2, st[5000], ed, lev[5000], des, res;
int sta[5000], stae[5000], stp, spre[5000];
struct node {
  int to, opt, next, fl, fr;
} g[600000];
inline void add(int x, int y, int z) {
  ed++;
  g[ed].fl = z;
  g[ed].to = y;
  g[ed].fr = x;
  g[ed].opt = ed + 1;
  g[ed].next = st[x];
  st[x] = ed;
  ed++;
  g[ed].fl = 0;
  g[ed].to = x;
  g[ed].fr = y;
  g[ed].next = st[y];
  g[ed].opt = ed - 1;
  st[y] = ed;
}
inline void extend() {
  int del, tem, temm;
  stp = 1;
  sta[1] = 0;
  for (int i = 0; i <= des; i++) {
    spre[i] = st[i];
  }
  while (stp != 0) {
    int tem = sta[stp];
    if (tem != des) {
      for (; spre[tem] != 0; spre[tem] = g[spre[tem]].next) {
        if (g[spre[tem]].fl != 0 && lev[tem] + 1 == lev[g[spre[tem]].to]) {
          break;
        }
      }
      if (spre[tem] != 0) {
        stp++;
        stae[stp] = spre[tem];
        sta[stp] = g[spre[tem]].to;
      } else {
        stp--;
        lev[tem] = -1;
      }
    } else {
      del = 10000000;
      for (int i = stp; i >= 2; i--) {
        del = min(g[stae[i]].fl, del);
      }
      res += del;
      for (int i = stp; i >= 2; i--) {
        g[stae[i]].fl -= del;
        g[g[stae[i]].opt].fl += del;
        if (g[stae[i]].fl == 0) stp = i - 1;
      }
    }
  }
}
inline bool makelevel() {
  queue<int> q;
  q.push(0);
  memset(lev, -1, sizeof(lev));
  lev[0] = 0;
  while (!q.empty()) {
    int tem = q.front();
    q.pop();
    for (int i = st[tem]; i != 0; i = g[i].next) {
      if (g[i].fl != 0 && lev[g[i].to] == -1) {
        lev[g[i].to] = lev[tem] + 1, q.push(g[i].to);
        if (g[i].to == des) return true;
      }
    }
  }
  return false;
}
int dinic() {
  res = 0;
  while (makelevel()) {
    extend();
  }
  return res;
}
inline bool isprime(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
int a[12000];
inline void makemap() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  des = n + 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 == 0)
      add(0, i, 2);
    else
      add(i, des, 2);
    for (int j = 1; j <= n; j++) {
      if (a[i] % 2 == 0 && a[j] % 2 == 1 && isprime(a[i] + a[j])) add(i, j, 1);
    }
  }
}
int vv[12000];
vector<int> ans[12000];
inline void dfs(int x, int y) {
  ans[y].push_back(x);
  vv[x] = 1;
  for (int i = st[x]; i != 0; i = g[i].next) {
    if (vv[g[i].to] == 0) {
      if (a[x] % 2 == 0) {
        if (g[i].fl == 0 && g[i].to != 0) dfs(g[i].to, y);
      } else {
        if (g[i].fl == 1 && g[i].to != des) dfs(g[i].to, y);
      }
    }
  }
}
inline void getanswer() {
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (vv[i] == 0) cnt++, dfs(i, cnt);
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d ", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) {
      printf("%d ", ans[i][j]);
    }
    printf("\n");
  }
}
int main() {
  makemap();
  if (dinic() != n)
    printf("Impossible\n");
  else
    getanswer();
  return 0;
}