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)

#pragma GCC optimize("O3")
#pragma GCC target("arch=corei7-avx")

#include <iostream>
#include <cstdio>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cassert>
#include <cmath>
#include <queue>
#include <ctime>

using namespace std;

#define REP(i, n) for(int (i) = 0; (i) < (n); (i)++)
#define SZ(a) (int)(a).size()
#define ALL(a) (a).begin(), (a).end()

int ex[555][555];
int pref[555][555];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m, q;
    cin >> n >> m >> q;
    REP(i, m) {
        int L, R;
        cin >> L >> R;
        ex[L][R]++;
    }
    for (int i = 1; i <= n; i++) {
        pref[i][0] = 0;
        for (int j = 1; j <= n; j++) {
            pref[i][j] = pref[i][j - 1] + ex[i][j];
        }
    }
    REP(i, q) {
        int L, R;
        cin >> L >> R;
        int res = 0;
        for (int j = L; j <= R; j++) {
            res += pref[j][R] - pref[j][j - 1];
        }
        cout << res << "\n";
    }
    return 0;
}