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>
#include <math.h>
#include <string.h>

int main(){
  char s[ 101 ];
  int n, m, Q;
  int i, j;
  int c;
  scanf( "%d%d%d", &n, &m, &Q );
  int l[ 200000 ], r[ 200000 ], lr[ 200000 ];
  //int p[ 100000 ], q[ 100000 ];
  for ( i = 0; i < m; i++ ){
    scanf( "%d%d", &l[ i ], &r[ i ] );
    lr[ i ] = r[ i ] - l[ i ];
  }
  for ( i = 0; i < Q; i++ ){
    c = 0;
    int p, q;
    scanf( "%d%d", &p, &q );
    //puts("");
    if ( q - p == n - 1 ){
      printf( "%d\n", m );
      continue;
    }
    for ( j = 0; j < m; j++ ){
      if ( lr[ j ] > q - p ) continue;
      if ( p > l[ j ] ) continue;
      if ( r[ j ] > q ) continue;
      // if ( p <= l[ j ] ) {
      //   if ( r[ j ] <= q ){
      //     c++;
      //   }
      // }
      c++;
    }
    printf( "%d\n", c );
  }


  return 0;
}