One day, an ant called Alice came to an M*M chessboard. She wanted to go around all the grids. So she began to walk along the chessboard according to this way: (you can assume that her speed is one grid per second)

  At the first second, Alice was standing at (1,1). Firstly she went up for a grid, then a grid to the right, a grid downward. After that, she went a grid to the right, then two grids upward, and then two grids to the left…in a word, the path was like a snake.

  For example, her first 25 seconds went like this:

  ( the numbers in the grids stands for the time when she went into the grids)

5

4

3

2

1

1          2          3           4           5

At the 8^th second , she was at (2,3), and at 20^th second, she was at (5,4).

Your task is to decide where she was at a given time.

(you can assume that M is large enough)

Input

  Input file will contain several lines, and each line contains a number N(1<=N<=2*10^9), which stands for the time. The file will be ended with a line that contains a number 0.

Output

  For each input situation you should print a line with two numbers (x, y), the column and the row number, there must be only a space between them.

Sample Input

8

20

25

0

Sample Output

2 3

5 4

1 5

注：
本題目意思很好理解，就是以螺旋的形式數字從小到大，一圈一圈旋轉！輸入數字N,輸出N在螺旋圈中的座標！本題的關鍵就是找出規律，首先可以確定在一個正方形中找規律，經觀察發現，正方形中對角線上的元素可以表示為a*(a -1）+1，其中a為正方形的邊長！從而再確定所找的數位於邊長為奇數還是邊長為偶數的正方形的最外層邊界上！再更行判斷就可以得出結果了！！
以下是AC程式碼：
#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
    int n , q ,ans;
    while(scanf("%d",&n) == 1)
    {
                if(!n)break;
                q = (int) ceil(sqrt(n));//找出正方形的邊長 
                ans = q * (q - 1) + 1;//對角線上的最外層的元素 
                if(q & 1)//注意判斷大小 
                {
                     if(n >= ans)
                          printf("%d %d\n",q -(n - ans),q);
                     else
                          printf("%d %d\n",q,q - (ans - n));
                }
                else
                {
                    if(n >= ans)
                        printf("%d %d\n",q,q -(n - ans));
                    else
                        printf("%d %d\n",q -(ans - n),q);
                        
                }
    }
    return 0;
}