Codeforces#419 Karen and Coffee

题目：

 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

input

2 1 1
1 1
200000 200000
90 100

output

0

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分析：

给出n个区间，再给出一个k值——要求在上述区间内出现的最小次数，再给出q个询问区间，问每个区间内有几个数符合要求。

代码：

#include <cstdio>
#include <cstring>
const int N=200005;
int num[N],sum[N];
int n,k,q;
int main(){
    int a,b;
    memset(num, 0, sizeof(num));    //num[i]表示i出现的次数
    memset(sum, 0, sizeof(sum));    //sum[i]表示从1到i符合要求的总数
    scanf("%d%d%d",&n,&k,&q);
    for (int i=0; i<n; i++) {
        scanf("%d%d",&a,&b);
        num[a]++;num[b+1]--;//思路：将区间左端加一，区间右端后一个数减一
    }
    for (int i=1; i<N; i++) {
        num[i]+=num[i-1];
        if(num[i]>=k)   sum[i]=1;
        sum[i]+=sum[i-1];
    }
    for (int i=0; i<q; i++) {
        scanf("%d%d",&a,&b);
        printf("%d\n",sum[b]-sum[a-1]);
    }
    return 0;
}