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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>

int main(){  
  long N,M,Q,L[200001], R[200001], p[100001], q[100001];

  scanf("%ld", &N);
  scanf("%ld", &M);
  scanf("%ld", &Q);

  long Lmin, Lmax;
  long Rmin, Rmax;

  for(long i=0; i<M; i++){
    scanf("%ld", &L[i]);
    scanf("%ld", &R[i]);
    if(i==0||Lmin>L[i]) Lmin = L[i];
    if(i==0||Lmax<L[i]) Lmax = L[i];
    if(i==0||Rmin>R[i]) Rmin = R[i];
    if(i==0||Rmax<R[i]) Rmax = R[i];
  }
  
  long ans[100001];

  for(long i=0; i<Q; i++){
    scanf("%ld", &p[i]);
    scanf("%ld", &q[i]);
    if(Lmax<p[i] || q[i]<Rmin){
      ans[i] = 0;
    }
    else{
      long num = 0;
      for(long j=0; j<M; j++){
	if(p[i]<=L[j] && R[j]<=q[i]){
	  num++;
	}
      }
      ans[i] = num;
    }
  }

  for(long i=0; i<Q; i++){
    printf("%ld\n", ans[i]);
  }

  return 0;
}