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;
void read(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -f;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  x *= f;
}
void print(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (!x) return;
  print(x / 10), putchar(x % 10 + 48);
}
void write(int x) {
  if (!x)
    putchar('0');
  else
    print(x);
  putchar('\n');
}
const int maxn = 1e6 + 10;
const int inf = 1e9;
const double eps = 1e-8;
const int mod = 1e9 + 7;
int n, m, head[maxn], tot = 1, a[maxn], c1, c2, s, t, max_flow, id[maxn],
                      cur[maxn];
struct edge {
  int to, nxt, w;
} e[maxn << 1];
void add(int u, int v, int w) {
  e[++tot] = (edge){v, head[u], w}, head[u] = tot;
}
void ins(int u, int v, int w) { add(u, v, w), add(v, u, 0); }
int dis[maxn];
int bfs() {
  memset(dis, -1, (n + 3) * 4);
  queue<int> q;
  q.push(s);
  dis[s] = 0;
  for (int i = 1; i <= n + 2; i++) cur[i] = head[i];
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = e[i].nxt)
      if (dis[e[i].to] == -1 && e[i].w != 0) {
        dis[e[i].to] = dis[x] + 1;
        if (e[i].to == t) return 1;
        q.push(e[i].to);
      }
  }
  return 0;
}
int dfs(int x, int f) {
  if (x == t) return f;
  int used = 0;
  for (int &i = cur[x]; i; i = e[i].nxt)
    if (dis[e[i].to] == dis[x] + 1 && e[i].w) {
      int d = dfs(e[i].to, min(f, e[i].w));
      e[i].w -= d, e[i ^ 1].w += d, used += d, f -= d;
      if (!f) break;
    }
  if (!used) dis[x] = -1;
  return used;
}
void dinic() {
  while (bfs()) max_flow += dfs(s, inf);
}
int check(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return 0;
  return 1;
}
int rev[maxn], vis[maxn];
vector<int> w[maxn], res;
vector<vector<int> > ans;
void dfs(int x) {
  vis[x] = 1;
  res.push_back(x);
  for (auto v : w[x])
    if (!vis[v]) dfs(v);
}
int main() {
  read(n);
  s = n + 1, t = n + 2;
  c2 = n / 2;
  for (int i = 1; i <= n; i++) {
    read(a[i]);
    if (a[i] & 1)
      id[i] = ++c1, rev[c1] = i;
    else
      id[i] = ++c2, rev[c2] = i;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if ((a[i] & 1) && !(a[j] & 1) && check(a[i] + a[j])) ins(id[i], id[j], 1);
  for (int i = 1; i <= n / 2; i++) ins(s, i, 2), ins(i + n / 2, t, 2);
  dinic();
  if (max_flow != n) return puts("Impossible"), 0;
  for (int x = 1; x <= n / 2; x++)
    for (int i = head[x]; i; i = e[i].nxt)
      if (e[i].to != s && !e[i].w)
        w[rev[x]].push_back(rev[e[i].to]), w[rev[e[i].to]].push_back(rev[x]);
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i), ans.push_back(res), res.clear();
  write(ans.size());
  for (auto r : ans) {
    printf("%d ", r.size());
    for (auto x : r) printf("%d ", x);
    puts("");
  }
  return 0;
}
