B. Lasers - 技术栈

time limit per test

2 seconds

memory limit per test

256 megabytes

There is a 2D-coordinate plane that ranges from (0,0) to (x,y). You are located at (0,0) and want to head to (x,y).

However, there are n horizontal lasers, with the i-th laser continuously spanning (0,ai) to (x,ai). Additionally, there are also m vertical lasers, with the i-th laser continuously spanning (bi,0) to (bi,y).

You may move in any direction to reach (x,y), but your movement must be a continuous curve that lies inside the plane. Every time you cross a vertical or a horizontal laser, it counts as one crossing. Particularly, if you pass through an intersection point between two lasers, it counts as two crossings.

For example, if x=y=2, n=m=1, a=[1], b=[1], the movement can be as follows:

What is the minimum number of crossings necessary to reach (x,y)?

Input

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

The first line of each test case contains four integers n, m, x, and y (1≤n,m≤2⋅105,2≤x,y≤109).

The following line contains n integers a1,a2,...,an (0<ai<y) --- the y-coordinates of the horizontal lasers. It is guaranteed that ai>ai−1 for all i>1.

The following line contains m integers b1,b2,...,bm (0<bi<x) --- the x-coordinates of the vertical lasers. It is guaranteed that bi>bi−1 for all i>1.

It is guaranteed that the sum of n and m over all test cases does not exceed 2⋅105.

Output

For each test case, output the minimum number of crossings necessary to reach (x,y).

Example

Input

Copy

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1 1 2 2

2 1 100000 100000

42 58

Output

Copy

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2
3

解题说明：此题是一道数学题，找规律能发现直接输出m+n即可。

cpp 复制代码

#include <stdio.h>

int main() 
{
	int t, n, m, x, y, a, ans;
	scanf("%d", &t);
	while (t--) 
	{
		scanf("%d %d %d %d", &n, &m, &x, &y);
		ans = n + m;
		while (n--)
		{
			scanf("%d", &a);
		}
		while (m--)
		{
			scanf("%d", &a);
		}
		printf("%d\n", ans);
	}
	return 0;
}