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>

typedef struct {
	int L;
	int R;
}pair;

int compair(const void *a, const void *b){
	int sub = ((pair *)a)->L - ((pair *)b)->L;
	if(sub != 0){
		return sub;
	}
	else{
		return ((pair *)a)->R - ((pair *)b)->R;
	}
}

int main(){
	int N, M, Q, p, q, i, j, k;
	scanf("%d%d%d", &N, &M, &Q);
	pair *ps = (pair *)malloc(sizeof(pair) * M);
	for(i = 0; i < M; i++){
		scanf("%d%d", &ps[i].L, &ps[i].R);
	}
	qsort(ps, M, sizeof(pair), compair);
	int **sum = (int **)malloc(sizeof(int *) * (N + 1));
	for(i = 0; i <= N; i++){
		sum[i] = (int *)malloc(sizeof(int) * (N + 1));
		for(j = 0; j <= N; j++){
			sum[i][j] = 0;
		}
	}
	for(i = N - 1, j = M - 1; i >= 0; i--){
		while(j >= 0){
			if(ps[j].L - 1 < i){
				break;
			}
			sum[i][ps[j].R]++;
			j--;
		}
		for(k = 1; k <= N; k++){
			sum[i][k] += sum[i][k - 1];
		}
		for(k = 1; k <= N; k++){
			sum[i][k] += sum[i + 1][k];
		}
	}
/*	for(i = 0; i < N; i++){
		for(j = 0; j <= N; j++){
			printf("%d ", sum[i][j]);
		}
		printf("\n");
	}
*/	for(i = 0; i < Q; i++){
		scanf("%d%d", &p, &q);
		printf("%d\n", sum[p - 1][q]);
	}
	return 0;
}