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;
const int inf = 1e9;
const int maxn = 210;
const int N = 1e5;
struct edge {
  int v, next, f;
} e[maxn * maxn << 1];
int h[maxn], hh[maxn], num = 1;
int n, src, sink;
bool f[N];
vector<int> v[maxn];
int a[maxn], cnt;
void build(int u, int v, int f) {
  num++;
  e[num].v = v;
  e[num].f = f;
  e[num].next = h[u];
  h[u] = num;
  num++;
  e[num].v = u;
  e[num].f = 0;
  e[num].next = h[v];
  h[v] = num;
}
int q[maxn], head, tail;
bool vis[maxn];
int dis[maxn];
bool bfs() {
  int u, v;
  memset(dis, 0, sizeof(dis));
  memset(vis, 0, sizeof(vis));
  vis[src] = true;
  head = 0;
  tail = 1;
  q[0] = src;
  while (head != tail) {
    u = q[head++];
    for (int i = h[u]; i; i = e[i].next) {
      v = e[i].v;
      if (e[i].f && !vis[v]) {
        q[tail++] = v;
        dis[v] = dis[u] + 1;
        vis[v] = true;
      }
    }
  }
  return vis[sink];
}
int dfs(int u, int delta) {
  if (u == sink)
    return delta;
  else {
    int res = 0;
    for (int i = hh[u]; i && delta; i = e[i].next)
      if (e[i].f && dis[e[i].v] == dis[u] + 1) {
        int dd = dfs(e[i].v, min(e[i].f, delta));
        e[i].f -= dd;
        e[i ^ 1].f += dd;
        delta -= dd;
        res += dd;
        hh[u] = i;
      }
    return res;
  }
}
int maxflow() {
  int res = 0;
  while (bfs()) {
    for (int i = 1; i <= sink; i++) hh[i] = h[i];
    res += dfs(src, inf);
  }
  return res;
}
void dfs2(int x) {
  vis[x] = true;
  v[cnt].push_back(x);
  for (int i = h[x]; i; i = e[i].next) {
    int v = e[i].v;
    if (vis[v] || e[i ^ (a[x] & 1)].f == 0 || v > n) continue;
    dfs2(v);
  }
}
int main() {
  for (int i = 2; i < N; i++)
    if (!f[i])
      for (int j = 2; i * j < N; j++) f[i * j] = true;
  scanf("%d", &n);
  src = n + 1;
  sink = n + 2;
  int cnt1 = 0, cnt2 = 0;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++)
    if (a[i] & 1)
      build(src, i, 2), cnt1++;
    else
      build(i, sink, 2), cnt2++;
  if (cnt1 != cnt2) return 0 * printf("Impossible\n");
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      if (!f[a[i] + a[j]]) {
        if (a[i] & 1)
          build(i, j, 1);
        else
          build(j, i, 1);
      }
  int ans = maxflow();
  if (ans != n) return 0 * printf("Impossible\n");
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      cnt++;
      dfs2(i);
    }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    int n = v[i].size();
    printf("%d", n);
    for (int j = 0; j < n; j++) printf(" %d", v[i][j]);
    printf("\n");
  }
  return 0;
}