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;
const int INF = 2e9;
struct edge {
  int last;
  int end;
  int weight;
} e[500005];
int n = 0, m = 0, s = 0, t = 0, a[205];
int ne = 1, note[1005];
int q[1005], d[1005];
bool vis[1005], pri[50005];
vector<int> r[205];
void NewEdge0(int u, int v, int w) {
  ne++;
  e[ne].last = note[u];
  e[ne].end = v;
  e[ne].weight = w;
  note[u] = ne;
}
void NewEdge(int u, int v, int w) {
  NewEdge0(u, v, w);
  NewEdge0(v, u, 0);
}
bool bfs() {
  for (int i = s; i <= t; i++) d[i] = 0;
  int head = 1, tail = 0;
  d[s] = 1, q[++tail] = s;
  while (head <= tail) {
    int u = q[head++];
    for (int i = note[u]; i; i = e[i].last)
      if (e[i].weight && !d[e[i].end]) {
        d[e[i].end] = d[u] + 1;
        if (e[i].end == t) return true;
        q[++tail] = e[i].end;
      }
  }
  return false;
}
int dfs(int x, int flow) {
  if (x == t) return flow;
  int res = flow;
  for (int i = note[x]; i; i = e[i].last)
    if (e[i].weight && d[e[i].end] == d[x] + 1) {
      int flw = dfs(e[i].end, min(e[i].weight, res));
      if (!flw) {
        d[e[i].end] = 0;
        continue;
      }
      e[i].weight -= flw;
      e[i ^ 1].weight += flw;
      res -= flw;
      if (!res) break;
    }
  return flow - res;
}
void dfs0(int x) {
  vis[x] = true;
  r[m].push_back((x <= n) ? x : (x - n));
  for (int i = note[x]; i; i = e[i].last) {
    if (vis[e[i].end]) continue;
    if (x <= n && ((i & 1) || e[i].weight)) continue;
    if (x > n && (!(i & 1) || !e[i].weight)) continue;
    dfs0(e[i].end);
  }
}
int main() {
  scanf("%d", &n);
  s = 0, t = n + n + 1;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  pri[1] = true;
  for (int i = 2; i <= 50000; i++)
    if (!pri[i])
      for (int j = i + i; j <= 50000; j += i) pri[j] = true;
  int stotflow = 0, ttotflow = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1)
      NewEdge(s, i, 2), stotflow += 2;
    else
      NewEdge(n + i, t, 2), ttotflow += 2;
    for (int j = i + 1; j <= n; j++) {
      if (pri[a[i] + a[j]]) continue;
      if (a[i] & 1)
        NewEdge(i, n + j, 1);
      else
        NewEdge(j, n + i, 1);
    }
  }
  int mxflow = 0;
  while (bfs())
    while (int fl = dfs(s, INF)) mxflow += fl;
  for (int i = 1; i <= n; i++)
    for (int j = note[i]; j; j = e[j].last)
      if ((j & 1) || e[j].weight) continue;
  if (stotflow != ttotflow) {
    printf("Impossible");
    return 0;
  }
  if (mxflow < stotflow)
    printf("Impossible");
  else {
    for (int i = 1; i <= n; i++)
      if ((a[i] & 1) && !vis[i]) m++, dfs0(i);
    printf("%d\n", m);
    for (int i = 1; i <= m; i++) {
      int len = r[i].size();
      printf("%d ", len);
      for (int j = 0; j < len; j++) printf("%d ", r[i][j]);
      printf("\n");
    }
  }
  return 0;
}
