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: Time Limit Exceeded)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
#include<vector>
#include<map>

using namespace std;
#define endl '\n'
#define pb push_back
#define fst first
#define scd second
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define MOD (1e9 + 7)
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define REP(i,n) for(int i = 0;i < (n);i++)
#define FOR(i,a,b) for(int i = (a);i <= (b);i++)
#define YES(n) cout << ((n) ? "YES" : "NO" ) << endl
#define Yes(n) cout << ((n) > "Yes" : "No" ) << endl


typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef vector<int> vi;
typedef vector<vector<int>> vii;

template <class T = int> T in() { T x; cin >> x; return (x); }
template <class T = int> void out(T x) { cout << (x) << endl; }

int gcd(int a, int b) { return b ? gcd(b, a%b) : a; }

void solve() {

	int n, m, q, bottom, top, ans, tmp1, tmp2;

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

	vector<vector<int>> t(n + 1);
	REP(i, n + 1) {
		t[i].resize(n + 1);
		REP(j, n + 1) {
			t[i][j] = 0;
		}
	}

	REP(i, m) {
		scanf("%d%d", &tmp1, &tmp2);
		t[tmp1][tmp2]++;
		//t[in()][in()]++;
	}

	REP(i, q) {
		//cin >> bottom >> top;
		scanf("%d%d", &bottom, &top);
		ans = 0;
		FOR(j, bottom, top) {
			FOR(k, bottom, top) {
				ans += t[j][k];
			}
		}
		//out(ans);
		printf("%d\n", ans);
	}


	return;
}

int main() {

	cin.tie(0);
	ios::sync_with_stdio(false);

	solve();

	return 0;

}