A. The Closest Pairtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputCurrently 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 dHere, 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?InputA single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).OutputIf 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. ExamplesInput4 3Output0 00 11 01 1Input2 100Outputno solution
Input4 3
Output0 00 11 01 1
['constructive algorithms', 'implementation', '*1300']