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: Accepted)

#include <stdio.h>

void prints(int n,int S[n+1][n+1] ){
  int i,j;
  for(i=1;i<n+1;i++){
    for(j=1;j<n+1;j++){
      printf("%d", S[j][i]);
    }
    printf("\n");
  }
  printf("\n");
}

int main(){
  int i,j,k,l,s,N,M,Q;
  scanf("%d %d %d", &N,&M,&Q);
  int L[M+1],R[M+1],p[Q],q[Q];
  for(i=1;i<M+1;i++){
    scanf("%d %d",&L[i],&R[i]);
  }
  for(i=0;i<Q;i++){
    scanf("%d %d", &p[i],&q[i]);
  }

  int S[N+1][N+1],Sum[N+1][N+1];
  for(i=0;i<N+1;i++){
    for(j=0;j<N+1;j++){
      S[i][j]=0;
      Sum[i][j]=0;
    }
  }

  for(i=1;i<M+1;i++){
    l=R[i];
    k=L[i];
    S[k][l]=S[k][l]+1;
  }
  //prints(N,S);
  
  for(i=1;i<N+1;i++){
    s=0;
    for(j=1;j<N+1;j++){
      Sum[i][j]=s+S[i][j];
      s=s+S[i][j];
    }
  }
  //prints(N,Sum);
  int ANS;
  for(i=0;i<Q;i++){
    ANS=0;
    for(k=p[i];k<N+1;k++){
      l=q[i];
      ANS=ANS+Sum[k][l];
    }
    printf("%d\n", ANS);
  }
  return 0;
}