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;
bool isp[20010];
void init() {
  for (int i = 2; i <= 20000; ++i) {
    int li = sqrt(i) + 0.1;
    isp[i] = 1;
    for (int j = 2; j <= li && isp[i]; ++j)
      if (i % j == 0) isp[i] = 0;
  }
}
const int maxn = 210, maxe = maxn * maxn * 3, oo = 1e9;
int e = 1, inx, vis[maxn], ln[maxn], ht[maxn], l, r, S, T;
int to[maxe], ne[maxe], he[maxn], cap[maxe];
int n, m;
int a[210];
int edge[210][210];
vector<int> ans[210];
vector<int> g[210];
void link(int x, int y, int v) {
  to[++e] = y, ne[e] = he[x], he[x] = e, cap[e] = 0;
  to[++e] = x, ne[e] = he[y], he[y] = e, cap[e] = v;
}
bool bfs() {
  for (ln[l = r = 1] = S, vis[S] = ++inx, ht[S] = 1; l <= r; l++)
    for (int j = he[ln[l]]; j; j = ne[j])
      if (vis[to[j]] < inx && cap[j ^ 1]) {
        vis[to[j]] = inx, ht[to[j]] = ht[ln[l]] + 1, ln[++r] = to[j];
        if (to[j] == T) return 1;
      }
  return 0;
}
int dfs(int f, int d) {
  if (!f || d == S) return f;
  int flow = 0, k;
  for (int j = he[d]; j; j = ne[j])
    if (vis[to[j]] == inx && ht[to[j]] == ht[d] - 1 &&
        (k = dfs(min(f, cap[j]), to[j]))) {
      f -= k, flow += k;
      cap[j] -= k, cap[j ^ 1] += k;
      if (!f) return flow;
    }
  ht[d] = maxn + 1;
  return flow;
}
int dinic() {
  int maxflow = 0;
  while (bfs()) maxflow += dfs(oo, T);
  return maxflow;
}
void dfs(int t) {
  vis[t] = 1;
  ans[m].push_back(t);
  for (int v : g[t])
    if (!vis[v]) dfs(v);
}
int main() {
  init();
  scanf("%d", &n);
  S = n + 1, T = S + 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    if (a[i] & 1)
      link(S, i, 2);
    else
      link(i, T, 2);
  }
  for (int i = 1; i <= n; ++i)
    if (a[i] & 1)
      for (int j = 1; j <= n; ++j)
        if (~a[j] & 1)
          if (isp[a[i] + a[j]]) {
            link(i, j, 1);
            edge[i][j] = e - 1;
          }
  int flow = dinic();
  if (flow == n) {
    for (int i = 1; i <= n; ++i)
      if (a[i] & 1)
        for (int j = 1; j <= n; ++j)
          if (~a[j] & 1)
            if (isp[a[i] + a[j]]) {
              if (cap[edge[i][j]]) g[i].push_back(j), g[j].push_back(i);
            }
    m = 0;
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; ++i)
      if (!vis[i]) {
        ans[m].clear();
        dfs(i);
        ++m;
      }
    printf("%d\n", m);
    for (int i = 0; i < m; ++i) {
      printf("%d", (int)ans[i].size());
      for (int v : ans[i]) printf(" %d", v);
      printf("\n");
    }
  } else
    printf("Impossible\n");
  return 0;
}