ACM (ACMers' Chatting Messenger) is a famous instant messaging software developed by Marjar Technology Company. To attract more users, Edward, the boss of Marjar Company, has recently added a new feature to the software. The new feature can be described as follows:

If two users, A and B, have been sending messages to each other on the last <var>m</var>consecutive days, the "friendship point" between them will be increased by 1 point.

More formally, if user A sent messages to user B on each day between the (<var>i</var> - <var>m</var> + 1)-th day and the <var>i</var>-th day (both inclusive), and user B also sent messages to user A on each day between the (<var>i</var> - <var>m</var> + 1)-th day and the <var>i</var>-th day (also both inclusive), the "friendship point" between A and B will be increased by 1 at the end of the <var>i</var>-th day.

Given the chatting logs of two users A and B during <var>n</var> consecutive days, what's the number of the friendship points between them at the end of the <var>n</var>-th day (given that the initial friendship point between them is 0)?

Input

There are multiple test cases. The first line of input contains an integer <var>T</var> (1 ≤ <var>T</var>≤ 10), indicating the number of test cases. For each test case:

The first line contains 4 integers <var>n</var> (1 ≤ <var>n</var> ≤ 109), <var>m</var> (1 ≤ <var>m</var> ≤ <var>n</var>), <var>x</var> and <var>y</var> (1 ≤ <var>x</var>, <var>y</var> ≤ 100). The meanings of <var>n</var> and <var>m</var> are described above, while <var>x</var> indicates the number of chatting logs about the messages sent by A to B, and <var>y</var> indicates the number of chatting logs about the messages sent by B to A.

For the following <var>x</var> lines, the <var>i</var>-th line contains 2 integers <var>l</var><var>a</var>, <var>i</var> and <var>r</var><var>a</var>, <var>i</var> (1 ≤ <var>l</var><var>a</var>,<var>i</var> ≤ <var>r</var><var>a</var>, <var>i</var> ≤ <var>n</var>), indicating that A sent messages to B on each day between the <var>l</var><var>a</var>, <var>i</var>-th day and the <var>r</var><var>a</var>, <var>i</var>-th day (both inclusive).

For the following <var>y</var> lines, the <var>i</var>-th line contains 2 integers <var>l</var><var>b</var>, <var>i</var> and <var>r</var><var>b</var>, <var>i</var> (1 ≤ <var>l</var><var>b</var>,<var>i</var> ≤ <var>r</var><var>b</var>, <var>i</var> ≤ <var>n</var>), indicating that B sent messages to A on each day between the <var>l</var><var>b</var>, <var>i</var>-th day and the <var>r</var><var>b</var>, <var>i</var>-th day (both inclusive).

It is guaranteed that for all 1 ≤ <var>i</var> < <var>x</var>, <var>r</var><var>a</var>, <var>i</var> + 1 < <var>l</var><var>a</var>, <var>i</var> + 1 and for all 1 ≤ <var>i</var> < <var>y</var>, <var>r</var><var>b</var>, <var>i</var> + 1 < <var>l</var><var>b</var>, <var>i</var> + 1.

Output

For each test case, output one line containing one integer, indicating the number of friendship points between A and B at the end of the <var>n</var>-th day.

Sample Input

2
10 3 3 2
1 3
5 8
10 10
1 8
10 10
5 3 1 1
1 2
4 5

Sample Output

3
0

Hint

For the first test case, user A and user B send messages to each other on the 1st, 2nd, 3rd, 5th, 6th, 7th, 8th and 10th day. As <var>m</var> = 3, the friendship points between them will be increased by 1 at the end of the 3rd, 7th and 8th day. So the answer is 3.

 

题意：

给出x(1<=x<=100)个区间和y(1<=y<=100)个区间，求出存在几个长度为m(1<=m<=n)公共子区间。

把题目样例看懂了，基本上题目就会做了。

// Asimple
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#define INF 0x3f3f3f3f
#define mod 2016
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 105;
int n, m, T, len, cnt, num, Max;
int x, y;

struct node{
    int l;
    int r;
};
node a[maxn], b[maxn];

void input() {
    scanf("%d", &T);
    while( T -- ) {
        cin >> n >> m >> x >> y;
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        for(int i=0; i<x; i++) {
            cin >> a[i].l >> a[i].r;
        }
        for(int i=0; i<y; i++) {
            cin >> b[i].l >> b[i].r;
        }
        int cnt = 0;
        for(int i=0; i<x; i++) {
            if( a[i].r-a[i].l+1 < m ) continue; 
            for(int j=0; j<y; j++) {
                if( b[j].r-b[j].l+1 < m ) continue;
                int l = max(a[i].l, b[j].l);
                int r = min(a[i].r, b[j].r);
                if( r - l + 1 >= m ) {
                    cnt += r-l+1 -m+1;
                }
            }
        }
        cout << cnt << endl;
    } 
}

int main() {
    input();
    return 0;
}