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;
inline int IN() {
  char ch = getchar();
  int f = 1, x = 0;
  while ((ch < '0' || ch > '9') && (ch != '-')) ch = getchar();
  if (ch == '-') f = -1, ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
  return x * f;
}
int n, a[100005], S, T, cnt;
int pr[100005];
struct edge {
  int u, v, w, pre;
} e[5000005];
int dex, adj[10005];
set<int> cho[100005];
vector<int> sol[100005];
bool V[100005];
void add(int x, int y, int w) {
  e[++dex] = (edge){x, y, w, adj[x]};
  adj[x] = dex;
  e[++dex] = (edge){y, x, 0, adj[y]};
  adj[y] = dex;
}
int Q[100005], l, r, dis[100005];
int bfs(int S, int T) {
  for (int i = (1); i <= (T); i++) dis[i] = -1;
  dis[Q[l = r = 1] = S] = 0;
  for (; l <= r; l++) {
    int u = Q[l];
    for (int i = adj[u]; i; i = e[i].pre) {
      if (e[i].w) {
        int v = e[i].v;
        if (dis[v] == -1) dis[Q[++r] = v] = dis[u] + 1;
        if (v == T) return 1;
      }
    }
  }
  return 0;
}
int dinic(int x, int f) {
  if (x == T) return f;
  int ret = 0;
  for (int i = adj[x]; i; i = e[i].pre) {
    int v = e[i].v;
    if (e[i].w && dis[v] == dis[x] + 1) {
      int flow = dinic(v, min(f - ret, e[i].w));
      e[i].w -= flow;
      e[i ^ 1].w += flow;
      ret += flow;
      if (ret == f) return ret;
    }
  }
  if (!ret) dis[x] = -1;
  return ret;
}
void dfs(int x) {
  if (V[x]) return;
  V[x] = 1;
  sol[cnt].push_back(x);
  int u = *cho[x].begin();
  cho[x].erase(u);
  cho[u].erase(x);
  dfs(u);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  S = n + 1;
  T = n + 2;
  dex = 1;
  for (int i = 2; i <= 50000; i++) {
    for (int j = i + i; j <= 50000; j += i) pr[j] = 1;
  }
  int sum1 = 0, sum2 = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1)
      add(S, i, 2), ++sum1;
    else
      add(i, T, 2), ++sum2;
  }
  if (sum2 != sum1) {
    puts("Impossible");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 == 0) continue;
    for (int j = 1; j <= n; j++) {
      if (a[j] % 2 == 1) continue;
      if (!pr[a[i] + a[j]]) add(i, j, 1);
    }
  }
  int ans = 0;
  while (bfs(S, T)) ans += dinic(S, 2100000000);
  if (ans != 2 * sum1) {
    puts("Impossible");
    return 0;
  }
  for (int i = (1); i <= (S - 1); i++) {
    if (a[i] % 2 == 0) continue;
    for (int x = adj[i]; x; x = e[x].pre) {
      if (e[x].v > n) continue;
      if (e[x].w == 0) cho[i].insert(e[x].v), cho[e[x].v].insert(i);
    }
  }
  for (int i = (1); i <= (S - 1); i++) {
    if (V[i]) continue;
    ++cnt;
    dfs(i);
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d ", sol[i].size());
    for (int j = 0; j < (int)sol[i].size(); j++) printf("%d ", sol[i][j]);
    puts("");
  }
  return 0;
}