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;
inline int gi() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum;
}
const int maxn = 205, maxa = 20005, Inf = 1 << 30;
int n, s, t, ans, a[maxn], isprm[maxa];
void pre() {
  static bool vis[maxa];
  static int p[maxa], pcnt;
  for (int i = 2; i < maxa; ++i) {
    if (!vis[i]) p[++pcnt] = i, isprm[i] = 1;
    for (int j = 1; j <= pcnt && p[j] * i < maxa; ++j) {
      vis[p[j] * i] = 1;
      if (i % p[j] == 0) break;
    }
  }
}
struct edge {
  int to, next, cap;
} e[maxn * maxn << 1];
int h[maxn], cur[maxn], tot = 1, vis[maxn], dis[maxn];
queue<int> que;
inline void add(int u, int v, int w) {
  e[++tot] = (edge){v, h[u], w};
  h[u] = tot;
  e[++tot] = (edge){u, h[v], 0};
  h[v] = tot;
}
bool bfs() {
  dis[s] = 0;
  vis[s] = 1;
  que.push(s);
  static int u;
  while (!que.empty()) {
    u = que.front();
    que.pop();
    for (int i = h[u], v; v = e[i].to, i; i = e[i].next)
      if (!vis[v] && e[i].cap) {
        vis[v] = 1;
        dis[v] = dis[u] + 1;
        que.push(v);
      }
  }
  return vis[t];
}
int dfs(int u, int a) {
  if (u == t || !a) return a;
  int f, flow = 0;
  for (int &i = cur[u], v; v = e[i].to, i; i = e[i].next)
    if (dis[v] == dis[u] + 1 && (f = dfs(v, min(e[i].cap, a)))) {
      e[i].cap -= f;
      e[i ^ 1].cap += f;
      flow += f;
      a -= f;
      if (!a) break;
    }
  return flow;
}
int max_flow() {
  int res = 0;
  while (bfs()) {
    memset(vis + 1, 0, sizeof(int) * t);
    memcpy(cur + 1, h + 1, sizeof(int) * t);
    res += dfs(s, Inf);
  }
  return res;
}
int m;
vector<int> to[maxn], vec[maxn];
void dfs(int u) {
  vec[m].push_back(u);
  vis[u] = 1;
  for (int v : to[u])
    if (!vis[v]) dfs(v);
}
int main() {
  n = gi();
  for (int i = 1; i <= n; ++i) a[i] = gi();
  pre();
  s = n + 1;
  t = s + 1;
  for (int i = 1; i <= n; ++i)
    if (a[i] & 1)
      add(s, i, 2), ans += 2;
    else
      add(i, t, 2);
  if (ans != n) return puts("Impossible"), 0;
  for (int i = 1; i <= n; ++i)
    if (a[i] & 1)
      for (int j = 1; j <= n; ++j)
        if (isprm[a[i] + a[j]]) add(i, j, 1);
  if (ans != max_flow()) return puts("Impossible"), 0;
  for (int u = 1; u <= n; ++u)
    if (a[u] & 1)
      for (int i = h[u], v; v = e[i].to, i; i = e[i].next)
        if (!e[i].cap && v <= n) to[u].push_back(v), to[v].push_back(u);
  memset(vis + 1, 0, sizeof(int) * t);
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) ++m, dfs(i);
  printf("%d\n", m);
  for (int i = 1; i <= m; ++i) {
    printf("%lu ", vec[i].size());
    for (int u : vec[i]) printf("%d ", u);
    puts("");
  }
  return 0;
}
