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> o, E, oi, ei;
int A[222];
int isP[30005];
bool vis[444];
const int INF = 100000000;
int dist[444];
int ptr[444];
int source = -1, sink = -1;
struct edge {
  int a, b, cap, flow;
};
vector<edge> e;
vector<int> g[444];
void addEdge(int a, int b, int c) {
  edge e1 = {a, b, c, 0};
  g[a].push_back((int)e.size());
  e.push_back(e1);
  edge e2 = {b, a, 0, 0};
  g[b].push_back((int)e.size());
  e.push_back(e2);
}
bool bfs() {
  memset(dist, -1, sizeof(dist));
  dist[source] = 0;
  queue<int> q;
  q.push(source);
  while (!q.empty() && dist[sink] == -1) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < g[u].size(); i++) {
      int id = g[u][i];
      int to = e[id].b;
      if (dist[to] != -1 || e[id].flow >= e[id].cap) continue;
      q.push(to);
      dist[to] = dist[u] + 1;
    }
  }
  return dist[sink] != -1;
}
int dfs(int u, int flow) {
  if (u == sink) return flow;
  if (!flow) return 0;
  for (; ptr[u] < g[u].size(); ptr[u]++) {
    int id = g[u][ptr[u]];
    int to = e[id].b;
    if (dist[to] != dist[u] + 1) continue;
    int pushed = dfs(to, min(flow, e[id].cap - e[id].flow));
    if (pushed) {
      e[id].flow += pushed;
      e[id ^ 1].flow -= pushed;
      return pushed;
    }
  }
  return 0;
}
int dinic() {
  int flow = 0;
  while (bfs()) {
    memset(ptr, 0, sizeof(ptr));
    int pushed;
    while ((pushed = dfs(source, INF))) {
      flow += pushed;
    }
  }
  return flow;
}
void pre() {
  for (int i = 2; i < 30000; i++) isP[i] = 1;
  for (int i = 2; i < 30000; i++)
    if (isP[i]) {
      for (int j = 2 * i; j < 30000; j += i) isP[j] = 0;
    }
}
int idx = 0;
vector<int> cycles[444];
int G[444][444];
void dfs2(int c, int n) {
  cycles[idx].push_back(c);
  for (int i = 1; i < n + 1; i++)
    if (!vis[i]) {
      if (G[c][i]) {
        vis[i] = true;
        dfs2(i, n);
      }
    }
}
int main() {
  int n;
  scanf("%d", &n);
  pre();
  for (int i = 1; i < n + 1; i++) {
    scanf("%d", &A[i]);
    if (A[i] & 1) {
      o.push_back(A[i]);
      oi.push_back(i);
    } else {
      E.push_back(A[i]);
      ei.push_back(i);
    }
  }
  if (o.size() != E.size()) {
    printf("Impossible\n");
    return 0;
  }
  n /= 2;
  source = 0;
  sink = 2 * n + 1;
  for (int i = 0; i < n; i++) {
    addEdge(source, i + 1, 2);
    addEdge(n + i + 1, sink, 2);
  }
  for (int i = 0; i < o.size(); i++)
    for (int j = 0; j < E.size(); j++) {
      if (isP[o[i] + E[j]]) {
        addEdge(i + 1, n + 1 + j, 1);
      }
    }
  int flow = dinic();
  if (flow != 2 * n) {
    printf("Impossible\n");
    return 0;
  }
  for (int i = 1; i < 2 * n + 1; i++) {
    for (int j = 0; j < g[i].size(); j++) {
      edge u = e[g[i][j]];
      if (u.flow) {
        G[i][u.b] = 1;
      }
    }
  }
  for (int i = 1; i < 2 * n + 1; i++)
    if (!vis[i]) {
      vis[i] = true;
      dfs2(i, 2 * n);
      idx++;
    }
  printf("%d\n", idx);
  for (int i = 0; i < idx; i++) {
    printf("%d", (int)cycles[i].size());
    for (int j = 0; j < cycles[i].size(); j++) {
      if (cycles[i][j] <= n)
        printf(" %d", oi[cycles[i][j] - 1]);
      else
        printf(" %d", ei[cycles[i][j] - n - 1]);
    }
    printf("\n");
  }
  return 0;
}