Problem:
Score: 400 points

Problem Statement

In Takahashi Kingdom, there is a east-west railroad and N cities along it, numbered 1, 2, 3, ..., N from west to east.
A company called AtCoder Express possesses M trains, and the train i runs from City L_i to City R_i (it is possible that L_i = R_i).
Takahashi the king is interested in the following Q matters:

The number of the trains that runs strictly within the section from City p_i to City q_i, that is, the number of trains j such that p_i \leq L_j and R_j \leq q_i.

Although he is genius, this is too much data to process by himself. Find the answer for each of these Q queries to help him.

Constraints

N is an integer between 1 and 500 (inclusive).

M is an integer between 1 and 200 \ 000 (inclusive).

Q is an integer between 1 and 100 \ 000 (inclusive).

1 \leq L_i \leq R_i \leq N (1 \leq i \leq M)

1 \leq p_i \leq q_i \leq N (1 \leq i \leq Q)

Input

Input is given from Standard Input in the following format:

N M Q
L_1 R_1
L_2 R_2
:
L_M R_M
p_1 q_1
p_2 q_2
:
p_Q q_Q

Output

Print Q lines. The i-th line should contain the number of the trains that runs strictly within the section from City p_i to City q_i.

Sample Input 1

2 3 1
1 1
1 2
2 2
1 2

Sample Output 1

3

As all the trains runs within the section from City 1 to City 2, the answer to the only query is 3.

Sample Input 2

10 3 2
1 5
2 8
7 10
1 7
3 10

Sample Output 2

1
1

The first query is on the section from City 1 to 7. There is only one train that runs strictly within that section: Train 1.
The second query is on the section from City 3 to 10. There is only one train that runs strictly within that section: Train 3.

Sample Input 3

10 10 10
1 6
2 9
4 5
4 7
4 7
5 8
6 6
6 7
7 9
10 10
1 8
1 9
1 10
2 8
2 9
2 10
3 8
3 9
3 10
1 10

Sample Output 3

7
9
10
6
8
9
6
7
8
10

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

//昇順
int asc(const void* a, const void* b){
    return *(int*)a-*(int*)b;
}

//降順
int desc(const void* a, const void* b){
    return *(int*)b-*(int*)a;
}

int main(void){
    int N,M,Q;
    scanf("%d%d%d",&N,&M,&Q);

    //map[L][R]は、Lから始まりRで終わる電車の数
    int map[N+1][N+1];
    for(int i=1;i<=N;i++){
        for(int j=1;j<=N;j++) map[i][j]=0;
    }
    //標準入力からmapの情報を入れる
    for(int i=0;i<M;i++){
        int l,r;
        scanf("%d %d",&l,&r);
        map[l][r]++;
    }

    int sum[N+1][N+1];
    //累積和をとる
    for(int i=1;i<=N;i++){
        sum[i][0]=0;
        sum[i][1]=map[i][1];
        for(int j=2;j<=N;j++){
            sum[i][j]=sum[i][j-1]+map[i][j];
        }
    }

    int p[Q];
    int q[Q];
    for(int i=0;i<Q;i++){
        scanf("%d%d",&p[i],&q[i]);
    }

    for(int k=0;k<Q;k++){
        int ans=0;
        for(int i=p[k];i<=q[k];i++){
            ans+=sum[i][q[k]]-sum[i][p[k]-1];
        }
        printf("%d\n",ans);
    }

    return 0;
}