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: Runtime Error)

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <functional>
#include <map>
#include <iomanip>
#include <math.h> 
#include <stack>
#include <queue>
#include <bitset>
#include <cstdlib>
#include <tuple>
#include <cctype>
#include <ctype.h>
#include <set>
#include <sstream>
#include <time.h>
#include <numeric>
using namespace std;
//#define int long long
#define rep(i,s,n) for(int i = s;i<n;i++)
#define repe(i,s,n) for(int i = s;i<=n;i++)
#define rrep(i,s,n) for(int i = (n)-1;i>=(s);i--)
#define all(v) (v).begin(),(v).end()
#define pb push_back
#define fi first
#define se second
#define chmin(a,b) a=min((a),(b))
#define chmax(a,b) a=max((a),(b))
#define l1 list[index]
#define l2 list[index - 1]
#define l3 list[index + 1]
#define iif(i,j) ((i<0 && j<0) || (i>0 && j>0)) ? true : false 
typedef long long ll;
typedef pair<int, int>pint;
typedef vector<int>vint;
typedef vector<pint>vpint;
typedef pair<pint, int> P1;
typedef pair<int, pint> P2;
typedef pair<pint, pint> PP;
static const ll maxLL = (ll)1 << 62;
const ll MOD = 1000000007;
const ll INF = 1e18;
const double PI = 3.14159265359;
int ca[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };



signed main() {
	int n, m, q, m2;
	vector<pair<int, int>>plist, qlist;
	int flag[502];

	memset(flag, 0, sizeof(flag));

	cin >> n >> m >> q;
	m2 = m;
	vector<vector<int>>list(502, vector<int>(2, 0)); // [0]は各個数、[1]はそれより大きい数値の数

	for (int i = 0; i < m; i++) {
		int l, r;
		cin >> l >> r;
		plist.push_back(make_pair(l, r));
		//flag[r]++;
		list[r][0]++;
	}

	for (int i = 0; i < q; i++) {
		int l, r;
		cin >> l >> r;
		qlist.push_back(make_pair(l, r));
	}

	sort(plist.begin(), plist.end());

	for (int i = 0; i < m; i++) {
		if (flag[plist[i].first] == 0) {
			flag[plist[i].first] = i + 1;
		}
	}

	for (int i = 500; i >= 0; i--) {
		if (flag[i - 1] == 0) {
			flag[i - 1] = flag[i];
		}
	}

	/*for (int i = 0; i <= 500; i++) {
		if (flag[i + 1] == 0) {
			flag[i + 1] = flag[i];
		}
	}*/

	int sum = 0;

	for (int i = 1; i <= 500; i++) {
		sum += list[i][0];
	}

	for (int i = 500; i >= 0; i--) {
		list[i][1] = sum;
		sum -= list[i][0];
	}

	vector<vector<vector<int>>>list2(200002);
	vector<vector<int>>moto = list;

	for (int i = 0; i < m; i++) {
		list2[i] = moto;
		int num = plist[i].second;
		moto[num][0]--;
		//for (int j = num + 1; j <= 500; j++) {
		for (int j = num ; j <= 500; j++) {
		//for (int j = 1; j <= num; j++) {
			if (moto[j][1] <= 0)break;
			moto[j][1]--;	
		}
	}

	//sort(qlist.begin(), qlist.end());
	int index = 0;

	for (int i = 0; i < q; i++) {
		int num = qlist[i].first;
		int num2 = qlist[i].second;

		cout << list2[flag[num] - 1][num2][1] << endl;

		//for (int j = index; j < m; j++) {
		//	if (plist[j].first >= num) {
		//		cout << list2[j][num2][1] << endl;
		//		//index = j;
		//		break;
		//	}
		//}
	}




	return 0;
}