C. The Legend of Freya the Frog

time limit per test

2 seconds

memory limit per test

256 megabytes

Freya the Frog is traveling on the 2D coordinate plane. She is currently at point (0,0)(0,0) and wants to go to point (x,y)(x,y). In one move, she chooses an integer dd such that 0≤d≤k0≤d≤k and jumps dd spots forward in the direction she is facing.

Initially, she is facing the positive xx direction. After every move, she will alternate between facing the positive xx direction and the positive yy direction (i.e., she will face the positive yy direction on her second move, the positive xx direction on her third move, and so on).

What is the minimum amount of moves she must perform to land on point (x,y)(x,y)?

Input

The first line contains an integer tt (1≤t≤1041≤t≤104) --- the number of test cases.

Each test case contains three integers xx, yy, and kk (0≤x,y≤109,1≤k≤1090≤x,y≤109,1≤k≤109).

Output

For each test case, output the number of jumps Freya needs to make on a new line.

Example

Input

Copy

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9 11 3

0 10 8

1000000 100000 10

Output

Copy

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8
4
199999

Note

In the first sample, one optimal set of moves is if Freya jumps in the following way: (0,00,0) →→ (2,02,0) →→ (2,22,2) →→ (3,23,2) →→ (3,53,5) →→ (6,56,5) →→ (6,86,8) →→ (9,89,8) →→ (9,119,11). This takes 8 jumps.

解题说明：此题是一道数学题，交替向上向右走，可以不走，请问到给定点需要走几次。由于可以走0步，所以向上走和向右走是相互独立的，只需要求出他们的最大值即可。注意走0步的移动也要统计在内。

cpp 复制代码

#include <iostream>
#include<algorithm>
using namespace std;

void solve()
{
	long long x, y, k;
	cin >> x >> y >> k;
	long long ansx = x / k + (x % k != 0), ansy = y / k + (y % k != 0);
	if (ansx > ansy)
	{
		cout << 2ll * ansx - 1 << endl;
	}
	else
	{
		cout << 2ll * ansy << endl;
	}
}
int main()
{
	int TT = 1;
	cin >> TT;
	while (TT--)
	{
		solve();
	}
	return 0;
}