976B Lara Croft and the New Game

B. Lara Croft and the New Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You might have heard about the next game in Lara Croft series coming out this year. You also might have watched its trailer. Though you definitely missed the main idea about its plot, so let me lift the veil of secrecy.

Lara is going to explore yet another dangerous dungeon. Game designers decided to use good old 2D environment. The dungeon can be represented as a rectangle matrix of n rows and m columns. Cell (x, y) is the cell in the x-th row in the y-th column. Lara can move between the neighbouring by side cells in all four directions.

Moreover, she has even chosen the path for herself to avoid all the traps. She enters the dungeon in cell (1, 1), that is top left corner of the matrix. Then she goes down all the way to cell (n, 1) — the bottom left corner. Then she starts moving in the snake fashion — all the way to the right, one cell up, then to the left to the cell in 2-nd column, one cell up. She moves until she runs out of non-visited cells. n and m given are such that she always end up in cell (1, 2).

Lara has already moved to a neighbouring cell k times. Can you determine her current position?

Input

The only line contains three integers n, m and k (2 ≤ n, m ≤ 109, n is always even, 0 ≤ k < n·m). Note that k doesn't fit into 32-bit integer type!

Output

Print the cell (the row and the column where the cell is situated) where Lara ends up after she moves k times.

Examples

input

Copy

4 3 0

output

Copy

1 1

input

Copy

4 3 11

output

Copy

1 2

input

Copy

4 3 7

output

Copy

3 2

Note

Here is her path on matrix 4 by 3:

The meaning of the question: an n*m proof, the starting point is 1, 1, then go down, go to the end and go to the right, and then go in a snake shape, which is the way of annotating the map, and then take k steps later , output the current position.

Solution: first judge, whether k is less than n, indicating that the number of steps taken can only be in the first column. The rest must go out of the first column, k minus the number of steps in the first column, is the number of steps in the remaining (m-1) column, use division and remainder operations, if the remaining x is odd, output nx, my. Otherwise, it is nx, y+2. This rule eg: 2 2 2 This group of samples was not released during the competition. After a few days, I searched blogs to see other people's solutions.

#include<bits/stdc++.h>
using namespace std;
intmain()
{
    long long n,m,k,x,y;
    cin>>n>>m>>k;
    if(k<n)
        cout<<k+1<<" "<<1;
    else
    {
        k-=n;
        x=k/(m-1);
        y=k%(m-1);
        if(x&1)
            cout<<n-x<<" "<<m-y;
        else cout<<n-x<<" "<<y+2;
    }
    return 0;
}