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 (x

The pseudo code of the unexpected code is as follows:

input n
for 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 second
d=INF        //here INF is a number big enough
tot=0
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

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

題解：說的是先按x再按y排序之後，跑完迴圈，讓tot超過k，操蛋啊，剛開始構造的x遞減，就納悶為啥一直錯，靠，

還排序。。。。

排序的話，直接讓所有x相等就行了，y不一樣就行。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+10;
int n,k;
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        if((n-1)*n/2<=k) printf("no solution\n");
        else
        {
            for(int i=1;i<=n;i++)
                printf("%d %d\n",0,i);
        }
    }
    return 0;
}