题干：

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 n
for i from 1 to ninput the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF        //here INF is a number big enough
tot=0
for i from 1 to nfor j from (i+1) to n++totif (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 print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.

The conditions below must be held:

• All the points must be distinct.
• |xi|, |yi| ≤ 109.
• After running the given code, the value of tot should be larger than k.

Examples

Input

4 3

Output

0 0
0 1
1 0
1 1

Input

2 100

Output

no solution

题目大意：

给你一个程序，要求让你出一组数据，使得这组数据会让程序的tot超过k。换句话说，给出一个最大循环次数k，并且已知如果tot超过k就会TLE。问：能不能给出个样例，让这代码TLE。

解题报告：

  先找到无论如何都不会让他TLE的情况，，也就是tot很小，而我们想让他TLE，，但是死活TLE不了，也就是我们想让tot尽量大，但是最大情况也是TLE不了的。。。也就是一个break也没有执行，那么肯定是tot最大的。。。所以我们就是构造一组解，让他一个break也执行不了，就最会让他TLE。就行了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
ll n,k;
int main()
{cin>>n>>k;if(n * (n-1) / 2 > k) {for(int i = 1; i<=n; i++) {printf("1 %d\n",i);}}else puts("no solution");return 0 ;}