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: Accepted)

#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

template<class T>
class Fenwick_tree{
	vector<T> a;
public:
	Fenwick_tree(int n):a(n){}
	void add(int i,T val){
		for(;i<a.size();i|=i+1) a[i]+=val;
	}
	T sum(int i,int j){
		if(i==0){
			T res{};
			for(j--;j>=0;j=(j&(j+1))-1) res+=a[j];
			return res;
		}
		return sum(0,j)-sum(0,i);
	}
};

int main(){
	int n,m,q; scanf("%d%d%d",&n,&m,&q);

	vector<pair<int,int>> Event[500];
	pair<int,int> I[200000],Q[100000];
	rep(i,m){
		scanf("%d%d",&I[i].first,&I[i].second);
		I[i].first--; I[i].second--;
		Event[I[i].first].emplace_back(0,i);
	}
	rep(i,q){
		scanf("%d%d",&Q[i].first,&Q[i].second);
		Q[i].first--; Q[i].second--;
		Event[Q[i].first].emplace_back(1,i);
	}

	int ans[100000];
	Fenwick_tree<int> F(n);
	for(int i=n-1;i>=0;i--) if(!Event[i].empty()) {
		for(auto a:Event[i]){
			int id=a.second;
			if(a.first==0){
				F.add(I[id].second,1);
			}
			else{
				ans[id]=F.sum(Q[id].first,Q[id].second+1);
			}
		}
	}

	rep(i,q) printf("%d\n",ans[i]);

	return 0;
}