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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>

void build_sum_table(int *L, int *R);
int get_sum_range(int L, int R);

int N, M, Q;
int sum[500 + 1][500 + 1] = {0};

int main (void)
{
    int L[200000], R[200000], p[100000], q[100000];
    int i;

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

    build_sum_table(L, R);

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

void build_sum_table(int *L, int *R)
{
    int i, j;

    for (i = 0; i < M; i++) {
        sum[L[i]][R[i]]++;
    }

    for (i = 1; i <= N; i++) {
        for (j = 1; j <= N; j++) {
            sum[i][j] += sum[i - 1][j];
            sum[i][j] += sum[i][j - 1];
            sum[i][j] -= sum[i - 1][j - 1];
        }
    }
}

int get_sum_range(int L, int R)
{
    int ans = sum[R][R] - sum[R][L - 1] - sum[L - 1][R] + sum[L - 1][L - 1];
    return ans;
}