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本文介绍了一个算法问题，涉及计算两个朋友在线聊天的时间段。通过输入各自不同时段的在线时间，算法确定在特定范围内有多少个整数时间点适合双方进行在线聊天。

Little X and Little Z are good friends. They always chat online. But both of them have schedules.

Little Z has fixed schedule. He always online at any moment of time between a₁ and b₁, betweena₂ andb₂, ..., betweena_p andb_p (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time0, he will be online at any moment of time between c₁ andd₁, betweenc₂ andd₂, ..., betweenc_q andd_q (all borders inclusive). But if he gets up at timet, these segments will be shifted byt. They become [c_i + t, d_i + t] (for alli).

If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time betweenl andr (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?

Input

The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).

Each of the next p lines contains two space-separated integersa_i, b_i (0 ≤ a_i < b_i ≤ 1000). Each of the next q lines contains two space-separated integersc_j, d_j (0 ≤ c_j < d_j ≤ 1000).

It's guaranteed that b_i < a_i + 1 andd_j < c_j + 1 for all validi and j.

Output

Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.

Sample test(s)

Input

1 1 0 4
2 3
0 1

Output

Input

Output

20

打表

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

int main(){
    int p,q,l,r,i,j,t,sum,flag,online,a,b;
    int c[60],d[60],time[2010];
    while(cin>>p>>q>>l>>r){
        sum=0;
        memset(time,0,sizeof(time));
        for(i=0;i<p;i++){
            cin>>a>>b;
            for(j=a;j<=b;j++){
                time[j]=1;
            }
        }
        for(i=0;i<q;i++){
            cin>>c[i]>>d[i];
        }
        for(i=l;i<=r;i++){
            flag=0;
            for(j=0;j<q;j++){
                for(online=c[j]+i;online<=d[j]+i;online++){
                    if(time[online]){
                        sum++;
                        flag=1;
                        break;
                    }
                }
                if(flag){
                    break;
                }
            }
        }
        cout<<sum<<endl;
    }
    return 0;
}