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;
vector<int> res[300];
int cas;
int pri[100000];
int a[300], p[300], q[300];
vector<int> M[300];
int n;
struct E {
  int v, w, k;
} e[40000];
int len;
vector<int> G[300];
int level[300];
int item[300];
void add(int u, int v, int w) {
  G[u].push_back(len);
  e[len].v = v, e[len].w = w, e[len].k = len + 1;
  len++;
  G[v].push_back(len);
  e[len].v = u, e[len].w = 0, e[len].k = len - 1;
  len++;
}
void init() {
  len = 0;
  for (int i = 0; i < 300; i++) G[i].clear();
}
bool bfs(int s, int t) {
  queue<int> q;
  while (!q.empty()) q.pop();
  q.push(s);
  memset(level, -1, sizeof(level));
  level[s] = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < G[u].size(); i++) {
      int k = G[u][i], v = e[k].v;
      if (level[v] == -1 && e[k].w) {
        level[v] = level[u] + 1;
        q.push(v);
      }
    }
  }
  if (level[t] == -1) return 0;
  return 1;
}
int dfs(int u, int t, int f) {
  if (u == t) return f;
  for (int &i = item[u]; i < G[u].size(); i++) {
    int k1 = G[u][i], v = e[k1].v, k2 = e[k1].k;
    if (level[v] > level[u] && e[k1].w) {
      int flow = dfs(v, t, min(f, e[k1].w));
      if (flow) {
        e[k1].w -= flow;
        e[k2].w += flow;
        return flow;
      }
    }
  }
  return 0;
}
int dinic(int s, int t) {
  int ans = 0;
  while (bfs(s, t)) {
    int flow;
    memset(item, 0, sizeof(item));
    while ((flow = dfs(s, t, 1000000000))) ans += flow;
  }
  return ans;
}
int vis[300];
void work(int s, int cur) {
  if (vis[s] && cur == s) return;
  vis[cur] = 1;
  res[cas].push_back(cur);
  for (int i = 0; i < M[cur].size(); i++) {
    if (!vis[M[cur][i]]) {
      work(s, M[cur][i]);
      return;
    }
  }
}
int main() {
  pri[5] = pri[2] = pri[3] = 1;
  for (int i = 6; i < 100000; i++) {
    int flag = 1;
    for (int j = 2; j < sqrt(i) + 1; j++) {
      if (i % j == 0) {
        flag = 0;
        break;
      }
    }
    if (flag) pri[i] = 1;
  }
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int temp = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 == 0)
      add(i, n + 1, 2), temp += 2;
    else
      add(0, i, 2);
    for (int j = i + 1; j <= n; j++) {
      if (pri[a[i] + a[j]] == 1) {
        if (a[i] % 2 == 0)
          add(j, i, 1);
        else
          add(i, j, 1);
      }
    }
  }
  int ans = dinic(0, n + 1);
  if (ans != temp) {
    printf("Impossible\n");
    return 0;
  }
  for (int u = 1; u <= n; u++)
    if (a[u] % 2 == 1) {
      for (int j = 0; j < G[u].size(); j++) {
        int k = G[u][j], v = e[k].v;
        if (e[k].w == 0) {
          M[u].push_back(v);
          M[v].push_back(u);
        }
      }
    }
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2 == 1 && !vis[i]) {
      work(i, i);
      cas++;
    }
  }
  printf("%d\n", cas);
  for (int i = 0; i < cas; i++) {
    printf("%d", res[i].size());
    for (int j = 0; j < res[i].size(); j++) {
      printf(" %d", res[i][j]);
    }
    printf("\n");
  }
}