Karen and Game

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.

Sample Input

Input

Output

Input

2 1 1
1 1
200000 200000
90 100

Output

Hint

In the first test case, Karen knows 3 recipes.

The first one recommends brewing the coffee between 91 and 94 degrees, inclusive.
The second one recommends brewing the coffee between 92 and 97 degrees, inclusive.
The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.

A temperature is admissible if at least 2 recipes recommend it.

She asks 4 questions.

In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.

In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.

In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.

In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.

In the second test case, Karen knows 2 recipes.

The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree.
The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.

A temperature is admissible if at least 1 recipe recommends it.

In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.

题目大意：

给你n个温度范围，如果里面某一个温度出现了不小于k次，那么这个温度就是可以接受的，现在经过q次询问，每次询问给你一个温度区间，让你说出里面有多少元素是符合要求的。

思路：

暴力肯定是不可能的，必然超时。那么用一个巧妙地方法来记录区间就可以。就是用一个数字来标记区间的开始与结束。。。。

就用1来表示区间开始，-1表示区间结束。

然后我们还需要引入前n项可接受数的和。

例如：食谱给出[10,90]区间，用刚刚的方法标记，就标记10这个温度为1，91这个温度为-1，所以，你从10跑到91这里面的和都是1，也就是都覆盖了一次。

当食谱再给一个空间[50,60]，我们还是用刚刚的方法进行一波标记，标记50为1,61为-1，这时从10跑的91，发现[10,50]的和都是1，[50,60]的和都是2，[61,90]的和都是1.

代码:

#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
using namespace std;
#define maxn 1000010
long long int data[maxn],dp[maxn];
int main()
{
    long long int m,n,i,j,k,h,t,l,r,q;
    while(scanf("%lld%lld%lld",&n,&m,&q)!=EOF)
    {
        memset(data,0,sizeof(data));
        memset(dp,0,sizeof(dp));
        for(i=0;i<n;i++)
        {
            scanf("%lld%lld",&l,&r);
            data[l]++,data[r+1]--;
        }
        k=0;
        for(i=1;i<=200000;i++)
        {
            k+=data[i];
            if(k>=m)
                dp[i]+=dp[i-1]+1;
            else
                dp[i]=dp[i-1];
        }
        for(i=0;i<q;i++)
        {
            scanf("%lld%lld",&l,&r);
            printf("%lld\n",dp[r]-dp[l-1]);
        }
    }
    return 0;
}

猜你喜欢