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 double PI = acos(0) * 2;
const double EPS = 1e-8;
const long long MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
const int oo = 1e9;
const double foo = 1e30;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcounll(s);
}
template <class T>
T sqr(T x) {
  return x * x;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
struct Dinic {
  static const int oo = 1e9;
  static const int MAXV = 1e5 + 10;
  static const int MAXE = 1e6 + 10;
  int n, s, t, E, adj[MAXE], next[MAXE], run[MAXV], last[MAXV], level[MAXV],
      que[MAXV];
  int flow[MAXE], cap[MAXE];
  void init(int _n, int _s, int _t) {
    n = _n;
    s = _s;
    t = _t;
    E = 0;
    fill_n(last, n + 1, -1);
  }
  void add(int u, int v, int c1, int c2) {
    adj[E] = v;
    flow[E] = 0;
    cap[E] = c1;
    next[E] = last[u];
    last[u] = E++;
    adj[E] = u;
    flow[E] = 0;
    cap[E] = c2;
    next[E] = last[v];
    last[v] = E++;
  }
  int bfs() {
    fill_n(level, n + 1, -1);
    level[s] = 0;
    int qsize = 0;
    que[qsize++] = s;
    for (int i = 0; i < qsize; i++) {
      for (int u = que[i], e = last[u]; ~e; e = next[e]) {
        int v = adj[e];
        if (flow[e] < cap[e] && !~level[v]) {
          level[v] = level[u] + 1;
          que[qsize++] = v;
        }
      }
    }
    return level[t] != -1;
  }
  int dfs(int u, int bot) {
    if (u == t) return bot;
    for (int& e = run[u]; ~e; e = next[e]) {
      int v = adj[e];
      int delta = 0;
      if (level[v] == level[u] + 1 && flow[e] < cap[e] &&
          (delta = dfs(v, min(bot, cap[e] - flow[e]))) > 0) {
        flow[e] += delta;
        flow[e ^ 1] -= delta;
        return delta;
      }
    }
    return 0;
  }
  int maxflow() {
    int total = 0;
    while (bfs()) {
      for (int i = 0; i <= n; i++) run[i] = last[i];
      for (int delta = dfs(s, oo); delta > 0; delta = dfs(s, oo)) {
        total += delta;
      }
    }
    return total;
  }
} dinic;
int m, vs[230], n, cnt[2], a[230], pr[20100];
vector<int> e[230], adj[230];
void Dfs(int u) {
  vs[u] = 1;
  e[m].push_back(u);
  for (int i = 0; i < ((int)(adj[u]).size()); i++) {
    int v = adj[u][i];
    if (vs[v]) continue;
    Dfs(v);
  }
}
int main() {
  pr[1] = 1;
  for (int i = 2; i < 2e4; i++) {
    if (!pr[i]) {
      for (int j = 2 * i; j < 2e4; j += i) {
        pr[j] = 1;
      }
    }
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    cnt[a[i] % 2]++;
  }
  if (cnt[0] != cnt[1]) {
    puts("Impossible");
    return 0;
  }
  dinic.init(n + 2, n, n + 1);
  for (int i = 0; i < n; i++) {
    if (a[i] % 2 == 0) {
      dinic.add(n, i, 2, 0);
    } else {
      dinic.add(i, n + 1, 2, 0);
    }
  }
  for (int i = 0; i < n; i++) {
    if (a[i] % 2 == 0) {
      for (int j = 0; j < n; j++) {
        if (a[j] % 2 && !pr[a[i] + a[j]]) {
          dinic.add(i, j, 1, 0);
        }
      }
    }
  }
  int fl = dinic.maxflow();
  if (fl != 2 * cnt[0]) {
    puts("Impossible");
    return 0;
  }
  for (int i = 0; i < dinic.E; i += 2) {
    if (dinic.flow[i] == 0) continue;
    int u = dinic.adj[i];
    int v = dinic.adj[i ^ 1];
    if (u != n && u != n + 1 && v != n && v != n + 1) {
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
  }
  for (int i = 0; i < n; i++) {
    if (!vs[i]) {
      Dfs(i);
      m++;
    }
  }
  printf("%d\n", m);
  for (int i = 0; i < m; i++) {
    printf("%d ", ((int)(e[i]).size()));
    for (int j = 0; j < ((int)(e[i]).size()); j++) {
      printf("%d ", e[i][j] + 1);
    }
    puts("");
  }
}