Codeforces Round #185 (Div. 2)-The Closest Pair

Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.

The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between(x₁, y₁) and(x₂, y₂) is.

The pseudo code of the unexpected code is as follows:

input nfor i from 1 to n    input the i-th point's coordinates into p[i]sort array p[] by increasing of x coordinate first and increasing of y coordinate secondd=INF        //here INF is a number big enoughtot=0for i from 1 to n    for j from (i+1) to n        ++tot        if (p[j].x-p[i].x>=d) then break    //notice that "break" is only to be                                            //out of the loop "for j"        d=min(d,distance(p[i],p[j]))output d

Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second,tot should not be more than k in order not to get Time Limit Exceeded.

You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?

Input

A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000,1 ≤ k ≤ 10⁹).

Output

If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else printn lines, and the i-th line contains two integersx_i, y_i(|x_i|, |y_i| ≤ 10⁹) representing the coordinates of thei-th point.

The conditions below must be held:

• All the points must be distinct.
• |x_i|, |y_i| ≤ 10⁹.
• After running the given code, the value of tot should be larger thank.

Example

Input

4 3

Output

0 00 11 01 1

Input

2 100

Output

no solution

题意理解：

找到让其超时的点，并输出n组超时的点，k是给定的最小时间。是给若是找不到，输出“No solution”

for i from 1 to n    for j from (i+1) to n        ++tot        if (p[j].x-p[i].x>=d) then break    //notice that "break" is only to be                                            //out of the loop "for j"        d=min(d,distance(p[i],p[j]))output d

从这可以看出tot的时间复杂度是n*(n-1)/2,这是最长的时间，但是如果两点的x坐标的距离超过d就会跳出。这时候就会跳出，用的时间会少。所以要让n*(n-1)/2>k,也就是最长的时间都超过k防止其不超过。

输出时的技巧：

d=INF!=0所以要输出点的横坐标都为0就可以！

以下为代码：

#include<iostream>#include<cstdio>using namespace std;int n,k;const int MAXN=2002;int A[MAXN];int main(){    scanf("%d%d",&n,&k);    double t=n*(n-1)/2;    if(t<=k) printf("no solution");    else{        for(int i=1;i<=n;i++){            cout<<'0'<<' '<<i<<endl;        }    }}