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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
#define REP(i, n) for(int i=0; i<n; i++)

int L[510];
int R[510];
ll SUM[510][510];


int N, M, Q;
void print(){
    for(int i = N-1; i >= 0; i--){
        for(int j = 0; j < N; j++){
            printf("%d ", SUM[j][i]);
        }
        printf("\n");
    }
}

int main(){
    cin >> N >> M >> Q;
    REP(i, M){
        int l, r;
        cin >> l >> r;
        l--, r--;
        L[i] = l;
        R[i] = r;
        SUM[r][l]++;
    }

    for(int r = 1; r < N; r++){
        SUM[r][0] += SUM[r-1][0];
    }
    for(int l = 1; l < N; l++){
        SUM[0][l] += SUM[0][l-1];
    }

    for(int r = 1; r < N; r++){
        for(int l = 1; l < N; l++){
            SUM[r][l] += SUM[r][l-1] + SUM[r-1][l] - SUM[r-1][l-1];
        }
    }

    //print();

    REP(i, Q){
        int p, q;
        cin >> p >> q;
        p--, q--;
        ll ans = 0;
        ans += SUM[q][q];
        if(p-1 >= 0){
            ans -= SUM[p-1][q];
            ans -= SUM[q][p-1];
            if(q-1 >= 0)
                ans += SUM[p-1][q-1];
        }
        cout << ans << endl;
    }
    return 0;
}