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;
inline char gc() {
  static char now[1 << 16], *S, *T;
  if (T == S) {
    T = (S = now) + fread(now, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline int read() {
  int x = 0;
  char ch = gc();
  while (ch < '0' || ch > '9') ch = gc();
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - '0';
    ch = gc();
  }
  return x;
}
struct node {
  int y, z, next, x;
} data[220 * 220];
int num = 1, prime[21000], h[220], T, n, level[220], age[220];
bool not_prime[21000], visit[220];
vector<int> print[220];
inline void insert1(int x, int y, int z) {
  data[++num].y = y;
  data[num].z = z;
  data[num].next = h[x];
  h[x] = num;
  data[num].x = x;
  data[++num].y = x;
  data[num].z = 0;
  data[num].next = h[y];
  h[y] = num;
  data[num].x = y;
}
inline bool bfs() {
  queue<int> q;
  memset(level, 0, sizeof(level));
  level[0] = 1;
  q.push(0);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = h[x]; i; i = data[i].next) {
      int y = data[i].y, z = data[i].z;
      if (level[y] || !z) continue;
      level[y] = level[x] + 1;
      q.push(y);
      if (y == T) return 1;
    }
  }
  return 0;
}
inline int dfs(int x, int s) {
  if (x == T) return s;
  int ss = s;
  for (int i = h[x]; i; i = data[i].next) {
    int y = data[i].y, z = data[i].z;
    if (level[x] + 1 == level[y] && z) {
      int xx = dfs(y, min(s, z));
      if (!x) level[y] = 0;
      s -= xx;
      data[i].z -= xx;
      data[i ^ 1].z += xx;
      if (!s) return ss;
    }
  }
  return ss - s;
}
inline void dfs1(int x, int id, int op) {
  print[id].push_back(x);
  visit[x] = 1;
  for (int i = h[x]; i; i = data[i].next) {
    int y = data[i].y, z = data[i].z;
    if (visit[y]) continue;
    if ((i ^ op) == 0) continue;
    if (z ^ op) dfs1(y, id, op ^ 1);
  }
}
int main() {
  n = read();
  T = n + 1;
  int odd = 0, even = 0;
  for (int i = 1; i <= n; ++i) {
    age[i] = read();
    if (age[i] % 2)
      insert1(0, i, 2), ++odd;
    else
      insert1(i, T, 2), ++even;
  }
  int tot = 0;
  if (odd != even) {
    printf("Impossible");
    return 0;
  }
  for (int i = 2; i <= 20000; ++i) {
    if (!not_prime[i]) prime[++tot] = i;
    for (int j = 1; prime[j] * i <= 2e4; ++j) {
      not_prime[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (age[i] % 2 == 0) continue;
    for (int j = 1; j <= n; ++j) {
      if (i == j) continue;
      int sum = age[i] + age[j];
      if (!not_prime[sum]) {
        insert1(i, j, 1);
      }
    }
  }
  int ans = 0;
  while (bfs()) ans += dfs(0, 0x3f3f3f3f);
  if (ans != n) {
    printf("Impossible");
    return 0;
  }
  tot = 0;
  visit[0] = 1;
  visit[T] = 1;
  for (int i = h[0]; i; i = data[i].next) {
    int y = data[i].y;
    if (!visit[y]) dfs1(y, ++tot, 1);
  }
  printf("%d\n", tot);
  for (int i = 1; i <= tot; ++i) {
    printf("%d ", print[i].size());
    for (int j = 0; j < print[i].size(); ++j) printf("%d ", print[i][j]);
    printf("\n");
  }
  return 0;
}