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 N = 409;
int n, a[N];
bool readInput() {
  scanf("%d", &n);
  int odd = 0, even = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] & 1)
      odd++;
    else
      even++;
  }
  if (odd != even) {
    return false;
  }
  return true;
}
const int NN = 3e4;
vector<bool> isPrime(NN + 1);
void sieve() {
  for (int i = 2; i <= NN; i++) {
    isPrime[i] = false;
  }
  for (int i = 3; i <= NN; i += 2) {
    isPrime[i] = true;
  }
  isPrime[2] = true;
  for (int i = 3; i * i <= NN; i += 2) {
    if (isPrime[i]) {
      for (int j = i * i; j <= NN; j += i) {
        isPrime[j] = false;
      }
    }
  }
}
int cap[N][N], deg[N], adj[N][N];
int q[N], preV[N];
int dinic(int n, int s, int t) {
  int flow = 0;
  while (true) {
    memset(preV, -1, sizeof(preV));
    int qf = 0, qb = 0;
    preV[q[qb++] = s] = -2;
    while (qb > qf && preV[t] == -1)
      for (int u = q[qf++], i = 0, v; i < deg[u]; i++)
        if (preV[v = adj[u][i]] == -1 && cap[u][v]) preV[q[qb++] = v] = u;
    if (preV[t] == -1) {
      break;
    }
    for (int z = 0; z < n; z++) {
      if (cap[z][t] && preV[z] != -1) {
        int bot = cap[z][t];
        for (int v = z, u = preV[v]; u >= 0; v = u, u = preV[v]) {
          bot = min(bot, cap[u][v]);
        }
        if (!bot) {
          continue;
        }
        cap[z][t] -= bot;
        cap[t][z] += bot;
        for (int v = z, u = preV[v]; u >= 0; v = u, u = preV[v]) {
          cap[u][v] -= bot;
          cap[v][u] += bot;
        }
        flow += bot;
      }
    }
  }
  return flow;
}
void initDinic() {
  memset(cap, 0, sizeof(cap));
  for (int i = 1; i <= n; i++) {
    if (a[i] & 1)
      cap[0][i] = 2;
    else
      cap[i][n + 1] = 2;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i != j) {
        if ((a[i] & 1) && !(a[j] & 1) && isPrime[a[i] + a[j]]) {
          cap[i][j] = 1;
        }
      }
    }
  }
  memset(deg, 0, sizeof(deg));
  for (int u = 0; u < n + 2; u++)
    for (int v = 0; v < n + 2; v++)
      if (cap[u][v] || cap[v][u]) adj[u][deg[u]++] = v;
}
bool used[N];
vector<int> g[N];
int main(int argc, char const *argv[]) {
  if (!readInput()) {
    printf("Impossible\n");
    return 0;
  }
  sieve();
  initDinic();
  if (dinic(n + 2, 0, n + 1) != n) {
    printf("Impossible\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if ((a[i] & 1) && !(a[j] & 1) && isPrime[a[i] + a[j]] && cap[i][j] == 0) {
        g[i].push_back(j);
        g[j].push_back(i);
      }
    }
  }
  vector<vector<int> > path;
  for (int i = 1; i <= n; ++i) {
    if (used[i]) {
      continue;
    }
    vector<int> current;
    int j = i;
    while (true) {
      current.push_back(j);
      used[j] = true;
      int nj = -1;
      for (int k = 0; k < g[j].size(); ++k) {
        if (!used[g[j][k]]) {
          nj = g[j][k];
          break;
        }
      }
      if (nj == -1) {
        break;
      }
      j = nj;
    }
    path.push_back(current);
  }
  printf("%d\n", path.size());
  for (int i = 0; i < path.size(); ++i) {
    printf("%d", path[i].size());
    for (int j = 0; j < path[i].size(); ++j) {
      printf(" %d", path[i][j]);
    }
    printf("\n");
  }
  return 0;
}
