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: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;


int main(){
    int N,M,Q;
    cin >> N >> M >> Q;
    vector<int>l(M),r(M),p(Q),q(Q),up(N),down(N),res(Q);
    for(int i=0;i<M;i++)cin >> l[i] >> r[i];
    for(int i=0;i<Q;i++)cin >> p[i] >> q[i];
    for(int i=0;i<N;i++){
        for(int j=0;j<M;j++){
            if(l[j]-1 == i)up[i]++;
            if(r[j]-1 == i)down[i]++;
        }
        if(i != 0){
            up[i] += up[i-1];
            down[i] += down[i-1];
        }
    }
    for(int i=0;i<Q;i++){
        if(p[i] == 1){
            if(up[q[i]-1] <= down[q[i]-1]){
                res[i] = up[q[i]-1];
            }
            else {
                res[i] = down[q[i]-1];
            }
        }
        else{
            if(up[q[i]-1] - up[p[i]-1]  <= down[q[i]-1] - down[p[i]-1] ){
                res[i] =  up[q[i]-1] - up[p[i]-1];
            } 
            else{
                res[i] =  down[q[i]-1] - down[p[i]-1];
            }
        }
    }
    for(int i=0;i<Q;i++)cout << res[i] << endl;
    return 0;
}