Fox And Dinner

Fox Ciel is participating in a party in Prime Kingdom. There are n foxes there (include Fox Ciel). The i-th fox is a _i 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 f ₁, f ₂, …, f _k are sitting around table in clockwise order, then for 1 ≤ i ≤ k - 1: f _i and f _i + 1 are adjacent, and f ₁ and f _k 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 a _i (2 ≤ a _i ≤ 10⁴).

Output

If it is impossible to do this, output “Impossible”.

Otherwise, in the first line output an integer m (): 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:

#include <cstring>
#include <vector>
#include  	<list>
#include

<map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>
#include <cassert>

using namespace std;

const int inf = (int)1e6;
const int N = 100010;

int fin;
int ss[N], ff[N], c[N], f[N], obr[N], pred[N], last[N], st[N];
int m;

void add(int x, int y, int z, int xx, int yy) {
m++;
ss[m] = x;
ff[m] = y;
c[m] = z;
f[m] = xx;
obr[m] = yy;
}

int x[N], d[N];

bool expath() {
for (int i = 0; i <= fin; i++) d[i] = -1;
  int b = 1, e = 1;
  x[1] = 0;
  d[0] = 0;
  while (b <= e) {
    int j = last[x[b]];
    while (j > 0) {
if (c[j] > f[j] && d[ff[j]] == -1) {
e++;
x[e] = ff[j];
d[x[e]] = d[x[b]] + 1;
if (x[e] == fin) {
return true;
}
}
j = pred[j];
}
b++;
}
return false;
}

int rec(int v, int w) {
if (v == fin) {
return w;
}
int r = 0, j = last[v];
while (j > 0) {
last[v] = pred[j];
if (c[j] > f[j] && d[ff[j]] == d[v] + 1) {
int ww = c[j] - f[j];
if (w - r < ww) ww = w - r;
      int t = rec(ff[j], ww);
      if (t > 0) {
f[j] += t;
if (obr[j] != 0) f[obr[j]] -= t;
r += t;
if (r == w) break;
}
}
j = pred[j];
}
return r;
}

bool prime[N];
bool was[N];
vector <int> nei[N];

int main() {
for (int q = 2; q < N; q++) {
    prime[q] = true;
  }
  for (int i = 2; i < N; i++) {
    if (prime[i]) {
      for (int j = i + i; j < N; j += i) {
        prime[j] = false;
      }
    }
  }
  int n;
  scanf("%d", &n);
  vector < pair <int, int> > even, odd;
for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    if (a % 2 == 0) {
      even.push_back(make_pair(a, i));
    } else {
      odd.push_back(make_pair(a, i));
    }
  }
  if (even.size() != odd.size()) {
    puts("Impossible");
    return 0;
  }
  int cnt = even.size();
  fin = cnt + cnt + 1;
  m = 0;
  for (int i = 0; i < cnt; i++) {
    add(0, i + 1, 2, 0, 0);
    add(cnt + 1 + i, fin, 2, 0, 0);
  }
  for (int i = 0; i < cnt; i++) {
    for (int j = 0; j < cnt; j++) {
      if (prime[even[i].first + odd[j].first]) {
        add(1 + i, 1 + cnt + j, 1, 0, m + 2);
        add(1 + cnt + j, 1 + i, 0, 0, m);
      }
    }
  }
  for (int i = 0; i <= fin; i++) {
    last[i] = 0;
  }
  for (int i = 1; i <= m; i++) {
    pred[i] = last[ss[i]];
    last[ss[i]] = i;
  }
  for (int i = 0; i <= fin; i++) {
    st[i] = last[i];
  }
  int ans = 0;
  while (expath()) {
    int t = rec(0, inf);
    ans += t;
    for (int i = 0; i <= fin; i++) {
      last[i] = st[i];
    }
  }
  if (ans != 2 * cnt) {
    puts("Impossible");
    return 0;
  }
  for (int i = 0; i < n; i++) {
    nei[i].clear();
  }
  for (int i = 1; i <= m; i++) {
    if (1 <= ss[i] && ss[i] <= cnt && cnt + 1 <= ff[i] && ff[i] <= cnt + cnt) {
      if (f[i] == 1) {
        int x = even[ss[i] - 1].second;
        int y = odd[ff[i] - cnt - 1].second;
        nei[x].push_back(y);
        nei[y].push_back(x);
      }
    }
  }
  vector < vector <int> > ret;
for (int i = 0; i < n; i++) {
    was[i] = false;
  }
  for (int i = 0; i < n; i++) {
    if (was[i]) {
      continue;
    }
    vector <int> add;
int cur = i;
int pr = nei[i][0];
while (true) {
add.push_back(cur);
was[cur] = true;
int to = nei[cur][0] + nei[cur][1] - pr;
pr = cur;
cur = to;
if (cur == i) {
break;
}
}
ret.push_back(add);
}
int sz = ret.size();
printf("%d\n", sz);
for (int i = 0; i < sz; i++) {
    int cc = ret[i].size();
    printf("%d", cc);
    for (int j = 0; j < cc; j++) {
      printf(" %d", ret[i][j] + 1);
    }
    printf("\n");
  }
  return 0;
}