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本文介绍了一种预测特殊棋盘上灯光亮起规律的方法。该棋盘为一个NxN的玻璃棋盘，每个单元格内都有一盏灯。文章提供了一个C++程序，能够根据输入的时间秒数预测哪个单元格的灯会亮起。该程序首先计算出灯亮起的周期规律，然后根据给定的时间确定具体的单元格位置。

H - 字

Time Limit:500MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

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Description

Fibsieve had a fantabulous (yes, it's an actual word) birthday party this year. He had so many gifts that he was actually thinking of not having a party next year.

Among these gifts there was an N x N glass chessboard that had a light in each of its cells. When the board was turned on a distinct cell would light up every second, and then go dark.

The cells would light up in the sequence shown in the diagram. Each cell is marked with the second in which it would light up.

(The numbers in the grids stand for the time when the corresponding cell lights up)

In the first second the light at cell (1, 1) would be on. And in the 5th second the cell (3, 1) would be on. Now, Fibsieve is trying to predict which cell will light up at a certain time (given in seconds). Assume that N is large enough.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case will contain an integer S (1 ≤ S ≤ 10¹⁵) which stands for the time.

Output

For each case you have to print the case number and two numbers (x, y), the column and the row number.

Sample Input

Sample Output

Case 1: 2 3

Case 2: 5 4

Case 3: 1 5

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
using namespace std;
LL n,m;
int main()
{
    LL y,x,k,t,s,l,r;
    LL cla;
    scanf("%lld",&cla);
    for(int gr=1; gr<=cla; gr++)
    {
        scanf("%lld",&n);
        printf("Case %lld: ",gr);
        s=sqrt(n);
        if(s*s==n)
        {
            if(!(s%2) )
                printf("%lld 1\n",s);
            else
                printf("1 %lld\n",s);
        }
        else
        {
            k=s+1;
            x=k*k-n;
            if(k%2==0)
            {
                if(x<k)
                    printf("%lld %lld\n",k,x+1);
                else printf("%lld %lld\n",2*k-x-1,k);
            }
            else
            {
                if(x<k)
                printf("%lld %lld\n",x+1,k);
                else printf("%lld %lld\n",k,2*k-1-x);
            }
        }
    }
    return 0;
}