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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
using namespace std;

int sum[501][501] = {};

// 二次元累積和によるO(N^2)解法

int main(){ 
    int n, m, Q;
    scanf("%d %d %d\n", &n, &m, &Q);

    int l, r, p, q;
    for(int i = 0; i < m; i++){
        scanf("%d %d\n", &l, &r);
        sum[l][r]++;
    }

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

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

    return 0;
}