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>
using namespace std;
const int inf = 1 << 30;
const int N = 200005;
int s, t, n, m, id, cnt, gs1, gs2, gs, top = 1, ans, x, y, z, head[N], dis[N],
                                       cur[N], prim[N], a[N], b[N];
bool vis[N], v[N];
struct node {
  int too, next, zh;
} edge[N];
vector<int> g[N];
inline int read() {
  int ret = 0, f = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) {
    ret = ret * 10 + c - 48;
    c = getchar();
  }
  if (f) return -ret;
  return ret;
}
void add(int a, int b, int c) {
  edge[++top].too = b;
  edge[top].zh = c;
  edge[top].next = head[a];
  head[a] = top;
}
bool bfs() {
  memset(dis, 0x3f, sizeof(dis));
  memset(vis, false, sizeof(vis));
  queue<int> q;
  dis[s] = 0;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    vis[u] = false;
    for (int i = head[u]; i; i = edge[i].next) {
      int v = edge[i].too;
      if (dis[v] > dis[u] + 1 && edge[i].zh) {
        dis[v] = dis[u] + 1;
        if (!vis[v]) {
          vis[v] = true;
          q.push(v);
        }
      }
    }
  }
  return dis[t] != 0x3f3f3f3f;
}
int dfs(int u, int flow) {
  int temp = flow, rlow = 0;
  if (u == t) return flow;
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].too;
    if (edge[i].zh && dis[v] == dis[u] + 1) {
      rlow = dfs(v, min(temp, edge[i].zh));
      if (!rlow) dis[v] = 0;
      edge[i].zh -= rlow;
      edge[i ^ 1].zh += rlow;
      if (!(temp -= rlow)) break;
    }
  }
  return flow - temp;
}
int dinic() {
  int x, ans = 0;
  while (bfs()) {
    while (x = dfs(s, inf)) {
      ans = ans + x;
    }
  }
  return ans;
}
void work(int n) {
  v[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!v[i]) prim[++cnt] = i;
    for (int j = 1; j <= cnt; j++) {
      if (i * prim[j] > n) break;
      v[i * prim[j]] = 1;
      if (i % prim[j] == 0) break;
    }
  }
}
void find(int x) {
  if (x >= 1 && x <= n) g[gs].push_back(x);
  vis[x] = 1;
  for (int i = head[x]; i; i = edge[i].next) {
    int v = edge[i].too;
    if (vis[v]) continue;
    if (edge[i | 1].zh == 1) find(v);
  }
}
signed main() {
  work(20000);
  n = read();
  for (int i = 1; i <= n; i++) {
    int x = read();
    if (x % 2 == 1)
      gs1++;
    else
      gs2++;
    a[i] = x;
  }
  if (gs1 != gs2) {
    puts("Impossible");
    return 0;
  }
  s = 0;
  t = n + 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 == 1) {
      add(s, i, 2);
      add(i, s, 0);
      for (int j = 1; j <= n; j++)
        if (!v[a[i] + a[j]]) {
          add(i, j, 1);
          add(j, i, 0);
        }
    } else {
      add(i, t, 2);
      add(t, i, 0);
    }
  }
  if (dinic() != n) {
    puts("Impossible");
    return 0;
  }
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      gs++;
      find(i);
    }
  cout << gs << endl;
  for (int i = 1; i <= gs; i++) {
    printf("%d ", g[i].size());
    for (int j = 0; j < g[i].size(); j++) printf("%d ", g[i][j]);
    puts("");
  }
  return 0;
}
