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 = 2e2 + 10;
const int maxe = maxn * maxn;
int fir[maxn], arr[maxn], cnt_edge, n, s, t;
bool mp[maxn][maxn];
struct Edge {
  int from, to, cap, flow, nxt;
  Edge(int u = 0, int v = 0, int c = 0, int f = 0, int n = 0)
      : from(u), to(v), cap(c), flow(f), nxt(n) {}
} edges[maxe];
void init() {
  memset(fir, -1, sizeof(fir));
  cnt_edge = 0;
}
void add_edge(int u = 0, int v = 0, int c = 0, int f = 0) {
  edges[cnt_edge] = Edge(u, v, c, f, fir[u]);
  fir[u] = cnt_edge++;
}
bool is_prime(int v) {
  for (int i = 2; i <= (int)sqrt(v); i++) {
    if (v % i == 0) return false;
  }
  return true;
}
int d[maxn], fa[maxn];
int max_flow() {
  int ret = 0;
  for (;;) {
    queue<int> Q;
    Q.push(s);
    memset(fa, -1, sizeof(fa));
    memset(d, 0, sizeof(d));
    d[s] = 100;
    while (!Q.empty()) {
      int u = Q.front();
      Q.pop();
      for (int e = fir[u]; e != -1; e = edges[e].nxt)
        if (!d[edges[e].to]) {
          if (edges[e].cap > edges[e].flow) {
            d[edges[e].to] = min(edges[e].cap - edges[e].flow, d[u]);
            fa[edges[e].to] = e;
            Q.push(edges[e].to);
          }
        }
      if (d[t]) break;
    }
    if (!d[t]) break;
    ret += d[t];
    for (int e = fa[t]; e != -1; e = fa[edges[e].from]) {
      edges[e].flow += d[t];
      edges[e ^ 1].flow -= d[t];
    }
  }
  return ret;
}
bool vis[maxn];
vector<int> ans[maxn];
int ans_cnt;
void dfs(int u) {
  vis[u] = true;
  ans[ans_cnt].push_back(u);
  for (int v = 0; v < n; v++)
    if (!vis[v] && mp[u][v]) {
      dfs(v);
    }
}
int main() {
  scanf("%d", &n);
  init();
  s = n, t = n + 1;
  int test = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] & 1) {
      if (i == 5) test++;
      add_edge(s, i, 2, 0);
      add_edge(i, s, 0, 0);
    } else {
      add_edge(i, t, 2, 0);
      add_edge(t, i, 0, 0);
    }
    for (int j = 0; j < i; j++) {
      if (is_prime(arr[i] + arr[j])) {
        if (arr[i] & 1) {
          add_edge(i, j, 1, 0);
          add_edge(j, i, 0, 0);
        } else {
          add_edge(j, i, 1, 0);
          add_edge(i, j, 0, 0);
        }
      }
    }
  }
  if (max_flow() != n) {
    printf("Impossible\n");
  } else {
    for (int u = 0; u < n; u++) {
      for (int e = fir[u]; e != -1; e = edges[e].nxt)
        if (edges[e].to != t && edges[e].to != n) {
          int v = edges[e].to;
          if (edges[e].flow == 1) {
            mp[u][v] = mp[v][u] = true;
          }
        }
    }
    for (int i = 0; i < n; i++)
      if (!vis[i]) {
        dfs(i);
        ans_cnt++;
      }
    printf("%d\n", ans_cnt);
    for (int i = 0; i < ans_cnt; i++) {
      printf("%d", (int)ans[i].size());
      for (int j = 0; j < (int)ans[i].size(); j++) {
        printf(" %d", ans[i][j] + 1);
      }
      printf("\n");
    }
  }
  return 0;
}