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<bits/stdc++.h>
using namespace std;
const int N = 200005;
const int M = 100005;
struct aa{
	int s;
	int e;
	int chang;
} a[N];
struct bb{
	int s;
	int e;
	int chang;
} b[M];
int cmp(bb a,bb b){
	return a.s<b.s;
}
int my_lower_bound(aa *a,int n,int x){
	int left=0,right=n-1;
	int mark;
	while(left<right){
		int m=left+(right-left)/2;
		if(a[m].s>=x){
			right=m;
			mark=right;
		}
		else{
			left=m+1;
			mark=left;
		}
	}
	return mark;
}
int my_upper_bound(aa *a,int n,int x){
	int left=0,right=n-1;
	int mark;
	while(left<right){
		int m=left+(right-left)/2;
		if(a[m].s>x){
			right=m;
			mark=right;
		}
		else{
			left=m+1;
			mark=left;
		}
	}
	return mark;
}
int main(){
	int n,m,q;
	scanf("%d %d %d",&n,&m,&q);
	for(int i=0;i<m;i++){
		scanf("%d %d",&a[i].s,&a[i].e);
		a[i].chang=a[i].e-a[i].s;
	}
//	
    
	for(int i=0;i<q;i++){
		int ans=0;
		scanf("%d %d",&b[i].s,&b[i].e);
//		b[i].chang=b[i].e-b[i].s;
		for(int tt=my_lower_bound(a,m,b[i].s);tt<m;tt++){
		    if(b[i].e>=a[tt].e)
		        ans++;			
		}

		printf("%d\n",ans);	
	}
/*    sort(b,b+q,cmp);
	for(int i=0;i<q;i++){
		
		for(int j=0;j<m;j++){
			if(b[i].chang<a[j].chang) continue;
			if(b[i].s<=a[j].s && b[i].e>=a[tt].e)
			    ans++;
		}
		printf("%d\n",ans);	
	}
	*/
    return 0;
}