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>
bool vis[100000 + 3];
int pr[20000];
int h;
int a[300];
int n;
int endp[200000], next[200000], cap[200000];
int gra[300], cur[300], dist[300], prev[300];
int cnt[300];
int ecnt;
char buf[10000], *bend;
char out[10000], *oend;
void add_edge(int u, int v, int c) {
  endp[++ecnt] = v;
  cap[ecnt] = c;
  next[ecnt] = gra[u];
  gra[u] = ecnt;
  endp[++ecnt] = u;
  cap[ecnt] = 0;
  next[ecnt] = gra[v];
  gra[v] = ecnt;
}
int isap(int vcnt, int S_, int T_) {
  memmove(cur, gra, sizeof(gra));
  memset(cnt, 0, sizeof(cnt));
  memset(dist, 0, sizeof(dist));
  prev[S_] = S_;
  int ret = 0, top;
  int u = S_;
  while (dist[S_] < vcnt) {
    int i = cur[u];
    for (; ~i; i = next[i]) {
      int v = endp[i];
      if (cap[i] && dist[u] == dist[v] + 1) {
        prev[v] = cur[u] = i;
        u = v;
        break;
      }
    }
    if (!~i) {
      cur[u] = gra[u];
      top = vcnt;
      for (int i = gra[u]; ~i; i = next[i]) {
        int v = endp[i];
        if (cap[i] && dist[v] < top) {
          top = dist[v];
        }
      }
      ++cnt[top + 1];
      if (!--cnt[dist[u]]) {
        goto STOP;
      }
      dist[u] = top + 1;
      if (u != S_) {
        u = endp[prev[u] ^ 1];
      }
    }
    if (u == T_) {
      int arg = 1 << 29;
      while (u != S_) {
        if (cap[prev[u]] < arg) {
          arg = cap[prev[u]];
        }
        u = endp[prev[u] ^ 1];
      }
      ret += arg;
      while (u != T_) {
        cap[cur[u]] -= arg;
        cap[cur[u] ^ 1] += arg;
        u = endp[cur[u]];
      }
      u = S_;
    }
  }
STOP:
  return ret;
}
int find(int u, int p) {
  vis[u] = true;
  sprintf(bend, " %d", u + 1);
  bend += strlen(bend);
  *bend = 0;
  for (int i = gra[u]; ~i; i = next[i]) {
    int v = endp[i];
    if (v == p || vis[v]) {
      continue;
    }
    if ((!!cap[i]) ^ (a[u] & 1)) {
      return find(v, u) + 1;
    }
  }
  return 1;
}
bool is_pr(int val) {
  int l = 0, r = h - 1;
  while (r - l > 1) {
    int mid = (l + r) >> 1;
    if (pr[mid] < val) {
      l = mid;
    } else {
      r = mid;
    }
  }
  return pr[r] == val;
}
int main() {
  for (int i = 2; i < 100000; ++i) {
    if (!vis[i]) {
      pr[h++] = i;
    }
    for (int j = 0; j < h; ++j) {
      if (100000 / pr[j] < i) {
        break;
      }
      vis[pr[j] * i] = true;
      if (!(i % pr[j])) {
        break;
      }
    }
  }
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; ++i) {
      scanf("%d", a + i);
    }
    memset(gra, 255, sizeof(gra));
    ecnt = -1;
    for (int i = 0; i < n; ++i) {
      if (a[i] & 1) {
        add_edge(n, i, 2);
        for (int j = 0; j < n; ++j) {
          if (!(a[j] & 1) && is_pr(a[i] + a[j])) {
            add_edge(i, j, 1);
          }
        }
      } else {
        add_edge(i, n + 1, 2);
      }
    }
    int flow = isap(n + 2, n, n + 1);
    if (flow != n) {
      puts("Impossible");
    } else {
      oend = out;
      memset(vis, 0, sizeof(vis));
      int cnt = 0;
      for (int i = 0; i < n; ++i) {
        if (!vis[i]) {
          bend = buf;
          sprintf(oend, "%d%s\n", find(i, -1), buf);
          oend += strlen(oend);
          ++cnt;
        }
      }
      *(++oend) = 0;
      printf("%d\n%s", cnt, out);
    }
  }
  return 0;
}
