codeforces-1009D Relatively Prime Graph

题目链接：http://codeforces.com/problemset/problem/1009/D

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

Construct a relatively prime graph with nn vertices and mm 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 nn and mm (1≤n,m≤1051≤n,m≤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 ii-th of the next mm lines should contain the ii-th edge (vi,ui)(vi,ui) of the resulting graph (1≤vi,ui≤n,vi≠ui1≤vi,ui≤n,vi≠ui). For each pair (v,u)(v,u) there can be no more pairs (v,u)(v,u) or (u,v)(u,v). The vertices are numbered from 11 to nn.

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

题意大概就是，给出n，1到n的n个点，问你用m条边连接这n个点可不可行，若不可行，输出“impossible"，若可行，输出“possible”，并且给出m条边分别连接的点，限制：每条边连接的两个点必须是互质的。

开始以为什么特殊的数学方法，后来发现直接枚举1到n里的i，j就可以过。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5+5;
pair<int, int> ans[N];

int gcd(int a,int b)    //判断是否互质
{
    if(a<b)
        return gcd(b,a);
    while(b)
    {
        int t=b;
        b=a%b;
        a=t;
    }
    return a;
}

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

	if (m < n - 1)     //连接n个点，最少n-1条边
    {
		puts("Impossible");
		return 0;
	}

	int cur = 0;

    for(int i=1;i<=n && cur<m;i++)
        for(int j=i+1;j<=n;j++)
    {
        if(gcd(i,j)==1)
        {
            ans[++cur]=make_pair(i,j);
            if(cur==m)
                break;
        }
    }

	if (cur != m)        //如果所有的互质数的对数不到m，连不成
    {
		puts("Impossible");
		return 0;
	}

	puts("Possible");
	for(int i=1;i<=m;i++)
		printf("%d %d\n", ans[i].first, ans[i].second);

	return 0;
}