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.
#include <bits/stdc++.h>
const int N = 4e5 + 5, M = 2e4 + 5, INF = 1e9;
using namespace std;
int n, s, t, ans, cnt, num = 1;
int used[N], a[N], f[N], d[N], p[M], ok[M], now[N];
int head[N], nt[N << 1], to[N << 1], val[N << 1], from[N << 1];
vector<int> ver[N];
bool BFS() {
  for (int i = s; i <= t; ++i) d[i] = 0;
  queue<int> q;
  q.push(s);
  now[s] = head[s];
  d[s] = 1;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x], y = to[i]; i; i = nt[i], y = to[i]) {
      if (!val[i] || d[y]) continue;
      d[y] = d[x] + 1;
      now[y] = head[y];
      if (y == t) return true;
      q.push(y);
    }
  }
  return false;
}
int dinic(int p, int flow) {
  if (p == t) return flow;
  int i, y, res = flow;
  for (i = now[p], y = to[i]; i && res; i = nt[i], y = to[i]) {
    if (d[y] != d[p] + 1 || !val[i]) continue;
    int k = dinic(y, min(res, val[i]));
    if (!k) {
      d[y] = 0;
      continue;
    }
    val[i] -= k;
    val[i ^ 1] += k;
    res -= k;
  }
  now[p] = i;
  return flow - res;
}
void Getprime() {
  ok[1] = 1;
  for (int i = 2; i < M; ++i) {
    if (!ok[i]) p[++cnt] = i;
    for (int j = 1; j <= cnt && p[j] * i < M; ++j) {
      ok[p[j] * i] = 1;
      if (i % p[j] == 0) break;
    }
  }
}
int Find(int x) { return f[x] == x ? x : f[x] = Find(f[x]); }
void Link(int x, int y) {
  int fx = Find(x), fy = Find(y);
  if (fx == fy) return;
  f[fx] = fy;
}
void Add(int x, int y, int z) {
  ++num;
  nt[num] = head[x];
  head[x] = num;
  to[num] = y;
  val[num] = z;
  from[num] = x;
  ++num;
  nt[num] = head[y];
  head[y] = num;
  to[num] = x;
  val[num] = 0;
  from[num] = y;
}
void Get(int p, int group) {
  ver[group].push_back(p);
  used[p] = 1;
  for (int i = head[p]; i; i = nt[i])
    if (!val[i ^ (a[p] % 2 == 0)] && !used[to[i]]) Get(to[i], group);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), f[i] = i;
  Getprime();
  s = 0;
  t = n + 1;
  for (int i = 1; i <= n; ++i) {
    if (a[i] & 1)
      Add(s, i, 2);
    else {
      Add(i, t, 2);
      continue;
    }
    for (int j = 1; j <= n; ++j)
      if (!ok[a[i] + a[j]]) Add(i, j, 1);
  }
  int Maxflow = 0, flow;
  while (BFS())
    while (flow = dinic(s, INF)) Maxflow += flow;
  if (Maxflow != n) return 0 * puts("Impossible");
  for (int i = 2; i <= num; i += 2)
    if (!val[i]) Link(from[i], to[i]);
  used[s] = used[t] = 1;
  for (int i = 1; i <= n; ++i)
    if ((a[i] & 1) && !used[i]) ++ans, Get(i, ans);
  printf("%d\n", ans);
  for (int i = 1; i <= ans; ++i) {
    printf("%d ", ver[i].size());
    for (int j = 0; j != ver[i].size(); ++j) printf("%d ", ver[i][j]);
    putchar(10);
  }
  return 0;
}
