cf#419 Karen and Coffee 前缀和

贴一发题目~~当初做的时候没做出来后来补出来的，学习了前缀和的处理
前缀和就是对于每一个区间的元素个数做一个总结
再dp一下求出重复值，重复值大于那个次数就可以

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 l_i and r_i degrees, inclusive, to achieve the optimal taste.

Karen thinks that a temperature is admissible if at leastk 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 betweena 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), andq (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, thei-th line among these contains two integersl_i andr_i (1 ≤ l_i ≤ r_i ≤ 200000), describing that thei-th recipe suggests that the coffee be brewed betweenl_i andr_i degrees, inclusive.

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

Output

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

Examples

Input

3 2 491 9492 9797 9992 9493 9795 9690 100

Output

3304

Input

2 1 11 1200000 20000090 100

Output

0
#include<stdio.h>#include<math.h>#include<stdlib.h>#include<string.h>using namespace std;int main(){    int n,k,q;    while(~scanf("%d %d %d",&n,&k,&q))    {        int l,r;        int num[200005];        memset(num,0,sizeof(num));        for(int i=0;i<n;i++)        {          scanf("%d %d",&l,&r);          num[l]++;          num[r+1]--;        }        int c[200005];        memset(c,0,sizeof(c));        for(int i=1;i<=200000;i++)        {            num[i]+=num[i-1];            c[i]+=c[i-1]+(num[i]>=k);        }        for(int i=0;i<q;i++)        {            scanf("%d %d",&r,&l);            printf("%d\n",c[l]-c[r-1]);        }    }    return 0;}