Karen and Coffee

To stay woke and attentive during classes, Karen needs some coffee!

Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed “The Art of the Covfefe”.

She knows n coffee recipes. The i-th recipe suggests that coffee should be brewed between li and ri degrees, inclusive, to achieve the optimal taste.

Karen thinks that a temperature is admissible if at least k recipes recommend it.

Karen has a rather fickle mind, and so she asks q questions. In each question, given that she only wants to prepare coffee with a temperature between a and b, inclusive, can you tell her how many admissible integer temperatures fall within the range?

Input

The first line of input contains three integers, n, k (1 ≤ k ≤ n ≤ 200000), and q (1 ≤ q ≤ 200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.

The next n lines describe the recipes. Specifically, the i-th line among these contains two integers li and ri (1 ≤ li ≤ ri ≤ 200000), describing that the i-th recipe suggests that the coffee be brewed between li and ri degrees, inclusive.

The next q lines describe the questions. Each of these lines contains a and b, (1 ≤ a ≤ b ≤ 200000), describing that she wants to know the number of admissible integer temperatures between a and b degrees, inclusive.

Output

For each question, output a single integer on a line by itself, the number of admissible integer temperatures between a and b degrees, inclusive.

Examples

Input
3 2 4
91 94
92 97
97 99
92 94
93 97
95 96
90 100

Output
3
3
0
4

题目大意是：输入三个数n，k，q。接下来n行，每次输入一个闭区间[a,b]，表示这个闭区间中的数出现过一次。接下来q行，每次输入一个闭区间[c,d]，求在闭区间[c,d]中，出现次数大于等于k的数的个数。

传统方法暴力跑二重循环会超时，因此使用一点骚操作——前缀和。

#include<iostream>#include<string.h>#include<stdio.h>#define MAXN 200010using namespace std;int main(){    int n,k,q,i,j;    int rec[MAXN],ans[MAXN];    int askbegin,askend,be,en;    while(cin>>n>>k>>q){        memset(rec,0,sizeof(rec));        memset(ans,0,sizeof(ans));         for(i=1;i<=n;i++){            cin>>be>>en;            rec[be]++;   //记录开始             rec[en+1]--;  //记录结束，因为是闭区间，到下一个才没有因此这里是en+1        }        for(i=1;i<=MAXN-1;i++){            rec[i]=rec[i]+rec[i-1];//类前缀和，表示某个数出现了多少次（如果是t个新的开始就比前面的数多了t，如果是t个结束就比前面的数少了t）            if(rec[i]>=k) ans[i]=ans[i-1]+1;            else ans[i]=ans[i-1];//前缀和，表示从1到现在这个数一共有几个是满足条件的，如果条件为真又多了1，反之则和前面的数保持一样        }         for(i=1;i<=q;i++){            cin>>askbegin>>askend;            cout<<ans[askend]-ans[askbegin-1]<<endl;   //因为是闭区间，所以askbegin-1         }    }       return 0;}