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: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<limits.h>

void Sort(int A[], int N);

int main()
{
	int N = 0, M = 0, Q = 0;
	int* L = (int*)calloc( 200000, sizeof(int) );
	int* R = (int*)calloc( 200000, sizeof(int));

	int* p = (int*)calloc( 100000, sizeof(int));
	int* q = (int*)calloc( 100000, sizeof(int));

	scanf("%d%d%d", &N, &M, &Q);

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

	int ans[100000] = { 0 };

	for (int i = 0; i < Q; i++) {
		for (int j = 0; j < M; j++) {
			if (p[i] <= L[j]) {
				if (q[i] >= R[j]) {
					ans[i]++;
				}
			}
		}
	}

	for (int i = 0; i < Q; i++) {
		printf("%d\n", ans[i]);
	}


}

void Sort(int A[], int N) {
	for (int j = 0; j < N ; j++) {
		for (int i = 0; i < N - j; i++) {
			if (A[i] < A[i + 1]) {
				int tmp = A[i];
				A[i] = A[i + 1];
				A[i + 1] = tmp;
			}
		}
	}
}