Algorithm

D. Relatively Prime Graph
time limit per test
2 seconds
memory limit per test
256 megabytes
Problem link- https://codeforces.com/contest/1009/problem/D
input
standard input
output
standard output

Let's call an undirected graph G=(V,E)�=(�,�) relatively prime if and only if for each edge (v,u)∈E(�,�)∈�  GCD(v,u)=1���(�,�)=1 (the greatest common divisor of v� and u� is 11). If there is no edge between some pair of vertices v� and u� then the value of GCD(v,u)���(�,�) doesn't matter. The vertices are numbered from 11 to |V||�|.

Construct a relatively prime graph with n� vertices and m� edges such that it is connected and it contains neither self-loops nor multiple edges.

If there exists no valid graph with the given number of vertices and edges then output "Impossible".

If there are multiple answers then print any of them.

Input

The only line contains two integers n� and m� (1≤n,m≤1051≤�,�≤105) — the number of vertices and the number of edges.

Output

If there exists no valid graph with the given number of vertices and edges then output "Impossible".

Otherwise print the answer in the following format:

The first line should contain the word "Possible".

The i�-th of the next m� lines should contain the i�-th edge (vi,ui)(��,��) of the resulting graph (1≤vi,ui≤n,vi≠ui1≤��,��≤�,��≠��). For each pair (v,u)(�,�) there can be no more pairs (v,u)(�,�) or (u,v)(�,�). The vertices are numbered from 11 to n�.

If there are multiple answers then print any of them.

Examples
input
Copy
5 6
output
Copy
Possible
2 5
3 2
5 1
3 4
4 1
5 4
input
Copy
6 12
output
Copy
Impossible
Note

Here is the representation of the graph from the first example:

 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

int const N = 1e5 + 1;
int n, m, v;
pair<int, int> phi[N];
set<pair<int, int> > e;

void print(int x, int &cnt) {
  for(int i = 1; cnt > 0 && i <= n; ++i)
    if(i != x && __gcd(x, i) == 1) {
      if(e.count({min(x, i), max(x, i)}) == 1)
        continue;
      e.insert({min(x, i), max(x, i)});
      printf("%d %d\n", x, i), --cnt;
    }
}

int main() {
  scanf("%d %d", &n, &m);

  for(int i = 1; i <= n; ++i)
    phi[i].first = phi[i].second = i;

  for(int p = 2; p <= n; ++p)
    if(phi[p].first == p) {
      phi[p].first = p - 1;

      for(int i = 2 * p; i <= n; i += p)
        phi[i].first = (phi[i].first / p) * (p - 1);
    }

  long long to = 0;
  for(int i = 1; i <= n; ++i)
    to += phi[i].first;

  if(to - 1 < m || m < n - 1) {
    puts("Impossible");
    return 0;
  }

  puts("Possible");

  for(int i = n; i > 0; --i)
    if(phi[i].first == phi[i].second - 1) {
      v = phi[i].second;
      break;
    }
  
  for(int i = 1; i <= n; ++i) {
    if(i == v)
      continue;

    printf("%d %d\n", v, i);
    e.insert({min(v, i), max(v, i)});
  }
  m -= (n - 1);

  sort(phi + 1, phi + n + 1);
  reverse(phi + 1, phi + n + 1);

  for(int i = 1; m > 0 && i <= n; ++i)
    print(phi[i].second, m);

  return 0;
}
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Input

x
–
+
cmd
5 6

Output

x
–
+
cmd
Possible
2 5
3 2
5 1
3 4
4 1
5 4
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Demonstration

Codeforces Solution-D. Relatively Prime Graph-Solution in C, C++, Java, Python,Relatively Prime Graph,Codeforces Solution

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