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;
const int MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
using namespace std;
struct E {
  int k, c;
  E() {}
  E(int _k, int _c) : k(_k), c(_c) {}
} es[4200010];
struct Flow {
  int n, m, dis[500010];
  int qq[500010], qr, ql;
  vector<int> e[500010];
  void init(int _n) {
    n = _n;
    m = 0;
    for (int i = 0; i < n; i++) e[i] = vector<int>();
  }
  void add(int a, int b, int c) {
    e[a].push_back(m);
    es[m] = E(b, c);
    m++;
    e[b].push_back(m);
    es[m] = E(a, 0);
    m++;
  }
  bool BFS() {
    memset(dis, -1, n * sizeof(int));
    ql = qr = 0;
    qq[qr++] = 0;
    dis[0] = 0;
    while (ql != qr && dis[n - 1] == -1) {
      int p = qq[ql++];
      for (__typeof((e[p]).begin()) it = (e[p]).begin(); it != (e[p]).end();
           it++) {
        E ee = es[*it];
        if (ee.c == 0 || dis[ee.k] != -1) continue;
        dis[ee.k] = dis[p] + 1;
        qq[qr++] = ee.k;
      }
    }
    return dis[n - 1] != -1;
  }
  long long go(int p, long long c) {
    if (p == n - 1) return c;
    long long ret = 0, tmp;
    for (__typeof((e[p]).begin()) it = (e[p]).begin(); it != (e[p]).end();
         it++) {
      E &ee = es[*it];
      if (ee.c == 0 || dis[p] + 1 != dis[ee.k]) continue;
      tmp = go(ee.k, min(c - ret, (long long)ee.c));
      ret += tmp;
      ee.c -= tmp;
      es[(*it) ^ 1].c += tmp;
      if (ret == c) break;
    }
    if (ret == 0) dis[p] = -1;
    return ret;
  }
  long long maxflow() {
    long long ret = 0;
    while (BFS()) ret += go(0, 1LL << 60);
    return ret;
  }
} flow;
int p[SIZE];
void prime_init() {
  for (int i = 2; i < SIZE; i++) {
    if (!p[i]) {
      for (int j = i + i; j < SIZE; j += i) {
        if (!p[j]) p[j] = i;
      }
    }
  }
}
bool is_prime(int x) {
  if (x == 1) return 1;
  return p[x] == x;
}
int a[222], gn, used[222];
vector<int> gg[222];
void dfs(int x) {
  if (used[x]) return;
  gg[gn].push_back(x);
  used[x] = 1;
  for (int i = 0; i < (((int)(flow.e[x]).size())); ++i) {
    int eid = flow.e[x][i];
    if (es[eid].k && es[eid].k != flow.n - 1) {
      if (a[x - 1] % 2 == 0) {
        if (es[eid].c == 0) dfs(es[eid].k);
      } else {
        if (es[eid].c == 1) dfs(es[eid].k);
      }
    }
  }
}
int main() {
  prime_init();
  int(n);
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &(a[i]));
  }
  flow.init(2 + n);
  for (int i = 0; i < (n); ++i) {
    if (a[i] % 2 == 0)
      flow.add(0, i + 1, 2);
    else
      flow.add(i + 1, n + 1, 2);
  }
  for (int i = 0; i < (n); ++i) {
    for (int j = 0; j < (n); ++j) {
      if (i == j) continue;
      if (!p[a[i] + a[j]]) {
        if (a[i] % 2 == 0) flow.add(i + 1, j + 1, 1);
      }
    }
  }
  if (flow.maxflow() != n)
    puts("Impossible");
  else {
    for (int i = 0; i < (n); ++i) {
      if (!used[i + 1]) {
        dfs(i + 1);
        gn++;
      }
    }
    printf("%d\n", gn);
    for (int i = 0; i < (gn); ++i) {
      printf("%d", ((int)(gg[i]).size()));
      for (int j = 0; j < (((int)(gg[i]).size())); ++j) printf(" %d", gg[i][j]);
      puts("");
    }
  }
  return 0;
}