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<iostream>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
int m1[200005][2],q1[100005][2];
void mergesort(int ns,int ne)
{
	if(ne-ns==1) 
		return;
	int m=ns+(ne-ns)/2;
	mergesort(ns,m);
	mergesort(m,ne);
	int ta[ne-ns][2];
	int nl=ns,nr=m;
	int nt=0;
	while(nl<m&&nr<ne)
	{
		if(m1[nl]<m1[nr]) {ta[nt][0]=m1[nl][0];ta[nt][1]=m1[nl][1];nl++;nt++;}
		else {ta[nt][0]=m1[nr][0];ta[nt][1]=m1[nr][1];nr++;nt++;}
	} 
	while(nl<m) {ta[nt][0]=m1[nl][0];ta[nt][1]=m1[nl][1];nl++;nt++;}
	while(nr<ne) {ta[nt][0]=m1[nr][0];ta[nt][1]=m1[nr][1];nr++;nt++;}
	for(int i=0;i<ne-ns;i++)
	{
		m1[i+ns][0]=ta[i][0];
		m1[i+ns][1]=ta[i][1];
	}
}
int main()
{
	int i,j,n,m,q;
	cin>>n>>m>>q;
	for(i=1;i<=m;i++)
	scanf("%d %d",&m1[i][0],&m1[i][1]);
	mergesort(1,m+1);
	for(i=1;i<=q;i++)
	{
		scanf("%d %d",&q1[i][0],&q1[i][1]);
		int left=1,right=m;
		while(left<right&&q1[i][0]>m1[left][0]){
			int mid=left+(right-left)/2;
			if(m1[mid][0]<q1[i][0])
			left=mid+1;
			else
			right=mid;
		}
		int count=0;
		while(m1[left][0]>=q1[i][0]&&left<=m)
		{
			if(m1[left][1]<=q1[i][1])
			count++;
			left++;
		}
		cout<<count<<"\n";
	}
	return 0;
}