D. Relatively Prime Graph

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

Construct a relatively prime graph with $\mathrm{nn vertices\; and mm edges\; such\; that\; it\; is\; connected\; and\; it\; contains\; neither\; self-loops\; nor\; multiple\; edges.}$

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 $\mathrm{nn and mm (1\le n,m\le 1051\le n,m\le 105)\; —\; the\; number\; of\; vertices\; and\; the\; number\; of\; edges.}$

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 $\mathrm{ii-th\; of\; the\; next mm lines\; should\; contain\; the ii-th\; edge (vi,ui)(vi,ui) of\; the\; resulting\; graph\; (1\le vi,ui\le n,vi\ne ui1\le vi,ui\le n,vi\ne ui).\; For\; each\; pair (v,u)(v,u)there\; can\; be\; no\; more\; pairs (v,u)(v,u) or (u,v)(u,v).\; The\; vertices\; are\; numbered\; from 11 to nn.}$

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

Note

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

没有想到正解居然是暴力。当时想n^2longn肯定过不了....

但是由于m是1e5，所以循环不到n^2次。

#include <bits/stdc++.h>
#define maxn 100005
using namespace std;
int main()
{
    vector<pair<int,int> > v;
    int cnt=0;
    int n,m;
    cin>>n>>m;
    if(m<n-1)
    {
        cout<<"Impossible"<<endl;
        return 0;
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            if(j!=i&&__gcd(i,j)==1)
            {
                v.push_back(make_pair(i,j));
                cnt++;
                if(cnt==m)
                {   cout<<"Possible"<<endl;
                    for(int i=0;i<v.size();i++)
        {
            cout<<v[i].first<<" "<<v[i].second<<endl;
        }
        return 0;
                }
            }
        }
    }
    cout<<"Impossible"<<endl;
    return 0;
}