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 maxn = 205;
const int maxv = 2e5 + 5;
const int inf = 1e9 + 7;
bool isprime[maxv];
void prework() {
  int n = 20000;
  isprime[1] = false;
  for (int i = 2; i <= n; i++) {
    int q = (int)sqrt(i);
    isprime[i] = true;
    for (int j = 2; j <= q; j++)
      if (i % j == 0) {
        isprime[i] = false;
        break;
      }
  }
}
int n;
int a[maxn];
int f[maxn][maxn];
queue<int> que;
int level[maxn];
bool bfs(int s, int t) {
  memset(level, -1, sizeof(level));
  level[s] = 0;
  que.push(s);
  while (!que.empty()) {
    int x = que.front();
    que.pop();
    for (int i = 1; i <= n + 1; i++) {
      if (level[i] == -1 && f[x][i] > 0) {
        level[i] = level[x] + 1;
        que.push(i);
      }
    }
  }
  if (level[t] == -1)
    return false;
  else
    return true;
}
int dfs(int x, int sum, int t) {
  if (x == t) return sum;
  int res = sum;
  for (int i = 1; i <= n + 1; i++) {
    if (sum == 0) break;
    if (f[x][i] > 0 && level[x] + 1 == level[i]) {
      int p = dfs(i, min(sum, f[x][i]), t);
      if (p > 0) {
        f[x][i] -= p;
        f[i][x] += p;
        sum -= p;
      }
    }
  }
  return res - sum;
}
int ans[maxn][maxn], ansn, ansc[maxn];
bool vis[maxn];
int main() {
  prework();
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= n; i++)
      for (int j = i + 1; j <= n; j++)
        if (isprime[a[i] + a[j]]) {
          if (a[i] & 1)
            f[i][j] = 1;
          else
            f[j][i] = 1;
        }
    for (int i = 1; i <= n; i++) {
      if (a[i] & 1)
        f[0][i] = 2;
      else
        f[i][n + 1] = 2;
    }
    int mf = 0;
    while (bfs(0, n + 1)) {
      mf += dfs(0, inf, n + 1);
    }
    if (mf != n) {
      printf("Impossible\n");
    } else {
      ansn = 0;
      memset(vis, false, sizeof(vis));
      for (int i = 1; i <= n; i++)
        if (!vis[i] && (a[i] & 1)) {
          ansn++;
          ansc[ansn] = 0;
          int now = i;
          while (1) {
            bool flag = true;
            ansc[ansn]++;
            vis[now] = true;
            ans[ansn][ansc[ansn]] = now;
            for (int j = 1; j <= n; j++)
              if (!vis[j] && f[j][now] == 1) {
                ansc[ansn]++;
                vis[j] = true;
                ans[ansn][ansc[ansn]] = j;
                for (int k = 1; k <= n; k++)
                  if (!vis[k] && f[j][k] == 1) {
                    now = k;
                    flag = false;
                    break;
                  }
                break;
              }
            if (flag) break;
          }
        }
      cout << ansn << endl;
      for (int i = 1; i <= ansn; i++) {
        printf("%d", ansc[i]);
        for (int j = 1; j <= ansc[i]; j++) printf(" %d", ans[i][j]);
        printf("\n");
      }
    }
  }
  return 0;
}