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;
struct edge {
  int b, l, c;
} g[50005 << 1];
int gap[205], cur[205], h[205], t, dis[205], n, pre[205], num[205];
int pri[200000 + 1], vis[200000 + 1];
vector<int> ans[205];
int arc(int i) { return i + (i & 1 ? 1 : -1); }
void addedge(int a, int b, int c) {
  g[++t] = (edge){b, h[a], c};
  h[a] = t;
  g[++t] = (edge){a, h[b], 0};
  h[b] = t;
}
void bfs(int u) {
  int i;
  queue<int> q;
  for (i = 0; i <= n; i++) dis[i] = n, gap[i] = 0;
  gap[dis[u] = 0] = 1;
  q.push(u);
  while (q.size()) {
    u = q.front();
    q.pop();
    for (i = h[u]; i; i = g[i].l)
      if (dis[g[i].b] >= n) {
        gap[dis[g[i].b] = dis[u] + 1]++;
        q.push(g[i].b);
      }
  }
}
int sap(int x, int y) {
  int flow = 0, i, d, u = x;
  bfs(y);
  for (i = 1; i <= n; i++) cur[i] = h[i];
  while (dis[x] < n) {
    for (i = cur[u]; i; i = g[i].l)
      if (dis[u] == dis[g[i].b] + 1 && g[i].c) break;
    if (i) {
      cur[u] = i;
      pre[u = g[i].b] = i;
      if (u == y) {
        for (i = pre[y], d = 1e8; i; i = pre[g[arc(i)].b]) d = min(d, g[i].c);
        for (flow += d, u = x, i = pre[y]; i; i = pre[g[arc(i)].b]) {
          g[i].c -= d;
          g[arc(i)].c += d;
        }
      }
    } else {
      for (i = cur[u] = h[u], d = n; i; i = g[i].l)
        if (g[i].c) d = min(d, dis[g[i].b] + 1);
      if (!--gap[dis[u]]) break;
      gap[dis[u] = d]++;
      if (u != x) u = g[arc(pre[u])].b;
    }
  }
  return flow;
}
void dfs(int u) {
  vis[u] = 1;
  ans[t].push_back(u - 2);
  for (int i = h[u]; i; i = g[i].l) {
    if ((i & 1) && !g[i].c && !vis[g[i].b]) dfs(g[i].b);
    if ((i & 1 ^ 1) && g[i].c && !vis[g[i].b]) dfs(g[i].b);
  }
}
int main() {
  int i, j;
  scanf("%d", &n);
  if (n & 1) {
    printf("Impossible");
    return 0;
  }
  for (i = 2; i <= 200000; i++) {
    if (!vis[i]) pri[++pri[0]] = i;
    for (j = 1; j <= pri[0] && i * pri[j] <= 200000; j++) {
      vis[i * pri[j]] = 1;
      if (!(i % pri[j])) break;
    }
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
    if (num[i] & 1) {
      addedge(1, i + 2, 2);
      for (j = 1; j < i; j++)
        if ((num[j] & 1 ^ 1) && !vis[num[i] + num[j]]) addedge(i + 2, j + 2, 1);
    } else {
      addedge(i + 2, 2, 2);
      for (j = 1; j < i; j++)
        if ((num[j] & 1) && !vis[num[i] + num[j]]) addedge(j + 2, i + 2, 1);
    }
  }
  n += 2;
  if (sap(1, 2) != n - 2)
    printf("Impossible");
  else {
    memset(vis, t = 0, (n + 1) * sizeof(int));
    for (i = vis[1] = vis[2] = 3; i <= n; i++)
      if (!vis[i]) ++t, dfs(i);
    printf("%d", t);
    for (i = 1; i <= t; i++) {
      printf("\n%d ", ans[i].size());
      for (j = 0; j < ans[i].size(); j++) printf("%d ", ans[i][j]);
    }
  }
  return 0;
}