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>
inline int read() {
  int data = 0, w = 1;
  char ch = getchar();
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
  return data * w;
}
const int N(205), LIM(20010), M(20000), INF(0x3f3f3f3f);
struct edge {
  int next, to, cap;
} e[N * N * 2];
int n, A[N], not_prime[LIM], head[N], e_num = -1, lev[N], cur[N], S, T;
inline void Add(int from, int to, int cap) {
  e[++e_num] = (edge){head[from], to, cap}, head[from] = e_num;
  e[++e_num] = (edge){head[to], from, 0}, head[to] = e_num;
}
void Init() {
  not_prime[1] = 1;
  for (int i = 2; i <= M; i++)
    if (!not_prime[i])
      for (int j = i * i; j <= M; j += i) not_prime[j] = 1;
}
int bfs() {
  std::queue<int> Q;
  Q.push(S);
  memset(lev, 0, (T + 1) << 2), lev[S] = 1;
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = head[x]; ~i; i = e[i].next) {
      int to = e[i].to;
      if (!e[i].cap || lev[to]) continue;
      lev[to] = lev[x] + 1, Q.push(to);
    }
  }
  return lev[T];
}
int dfs(int x, int f) {
  if (x == T || !f) return f;
  int ans = 0, cap;
  for (int i = head[x]; ~i; i = e[i].next) {
    int to = e[i].to;
    if (e[i].cap && lev[to] == lev[x] + 1) {
      cap = dfs(e[i].to, std::min(f - ans, e[i].cap));
      e[i].cap -= cap, e[i ^ 1].cap += cap, ans += cap;
      if (ans == f) break;
    }
  }
  if (!ans) lev[x] = 0;
  return ans;
}
int Dinic() {
  int ans = 0;
  while (bfs()) memcpy(cur, head, (T + 1) << 2), ans += dfs(S, INF);
  return ans;
}
std::vector<int> cir[N];
int tot, vis[N];
void find(int x) {
  if (1 <= x && x <= n) cir[tot].push_back(x);
  vis[x] = 1;
  for (int i = head[x]; ~i; i = e[i].next)
    if (!vis[e[i].to] && e[i | 1].cap == 1) find(e[i].to);
}
int main() {
  memset(head, -1, sizeof head);
  Init(), n = read(), S = 0, T = n + 1;
  int p = 0, q = 0;
  for (int i = 1; i <= n; i++) A[i] = read(), ((A[i] & 1) ? ++p : ++q);
  if (p != q) return puts("Impossible"), 0;
  for (int i = 1; i <= n; i++)
    if (A[i] & 1)
      for (int j = (Add(S, i, 2), 1); j <= n; j++) {
        if (!not_prime[A[i] + A[j]]) Add(i, j, 1);
      }
    else
      Add(i, T, 2);
  if (Dinic() != n) return puts("Impossible"), 0;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) ++tot, find(i);
  printf("%d\n", tot);
  for (int i = 1; i <= tot; i++, puts("")) {
    printf("%lu ", cir[i].size());
    for (int j : cir[i]) printf("%d ", j);
  }
  return 0;
}
