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<algorithm>
#include<cmath>
#include<iomanip>
#include<iostream>
#include<map>
#include<numeric>
#include<queue>
#include<set>
#include<sstream>
#include<unordered_map>
#include<unordered_set>
#include<vector>
using ll = long long;
enum : int { M = (int)1e9 + 7 };
enum : ll { MLL = (ll)1e18L + 9 };
using namespace std;
#ifdef LOCAL
#include"rprint2.hpp"
#include"debug_deque.hpp"
#define vector DebugDeque
#else
#define FUNC(name) template <ostream& out = cout, class... T> void name(T&&...){ }
FUNC(prints) FUNC(printe) FUNC(printw) FUNC(printew) FUNC(printb) FUNC(printd) FUNC(printde) FUNC(printdu);
#endif
template <class S, class T>
istream& operator >> (istream& in, pair<S, T>& p){ return in >> p.first >> p.second; }
template <class T>
istream& operator >> (istream& in, vector<T>& v){ for(auto& e : v){ in >> e; } return in; }

#include<unistd.h>
struct Buffer{
    enum { chunk = 1 << 16 };
    int idx;
    char buf[chunk];
    Buffer(){ init(); }
    void init(){ idx = 0; read(0, buf, chunk); }
    void check(){ if(idx == chunk){ init(); } }
    int next(){
        static char notDigits[256] = {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1, 0,0,0,0,0,0,0,0,0,0,};
        int ret = 0, sign = 1;
        while(check(), notDigits[(int)buf[idx]]){ idx++; }
        if(check(), buf[idx] == '-'){ sign = -1; idx++; }
        while(check(), !notDigits[(int)buf[idx]]){ ret = ret * 10 + buf[idx++] - '0'; }
        return sign * ret;
    }
};
struct FastInput {
    Buffer buf;
    template <class T>
    FastInput& operator >> (T& num) { num = buf.next(); return *this; }
    void tie(int){ }
    template <class S, class T>
    friend FastInput& operator >> (FastInput& in, pair<S, T>& p){ return in >> p.first >> p.second; }
    template <class T>
    friend FastInput& operator >> (FastInput& in, vector<T>& v){ for(auto& e : v){ in >> e; } return in; }
};
FastInput fin;
#define cin fin

int accs[500][500];
int main(){
    int n, m, q;
    cin >> n >> m >> q;
    // vector<vector<int>> accs(n, vector<int>(n));
    for(int i = 0; i < m; i++){
        int l, r;
        cin >> l >> r;
        l--; r--;
        for(int i = 0; i <= l; i++){
            accs[i][r]++;
        }
    }
    for(int i = 0; i < n; i++){
        for(int j = 1; j < n; j++){
            accs[i][j] += accs[i][j - 1];
        }
    }
    for(int i = 0; i < q; i++){
        int p, q;
        cin >> p >> q;
        p--; q--;
        cout << accs[p][q] << '\n';
    }
}