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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#include<math.h>
#include<list>
#include <algorithm>
#include<set>
#include<vector>
#include<math.h>
#include<map>
#include<string>
#include <numeric>
#include <queue>
#include <sstream>
#include <bitset>
#include<stack>
#include<functional>
#include<unordered_set>
#include<unordered_map>

using namespace std;
using ll = long long;
using vll = vector<long long>;
using sll = set<long long>;
const long long ll_max = 9223372036854775807;
const long long ll_min = -9223372036854775807;
const int int_max = 2147483647;

template<typename T>
T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<typename T>
T lcm(T a, T b) { return a / gcd(a, b) * b; }

template<typename T>
map<T, T> getPrimeFactor(T n) {
	map<T, T> ret;
	for (T i = 2; i * i <= n; ++i) {
		while (n % i == 0) {
			ret[i] ++;
			n /= i;
		}
	}
	if (n != 1) ret[n] = 1;
	return ret;
}

template<typename T>
bool IsPrimeNumber(T num)
{
	if (num <= 2) return true;
	else if (num % 2 == 0) return false;

	double sqrtNum = sqrt(num);
	for (int i = 3; i <= sqrtNum; i += 2)
	{
		if (num % i == 0)
		{
			return false;
		}
	}

	return true;
}

template<typename T>
T modinv(T a, T m) {
	T b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}



// 繰り返し二乗法
ll pow2(ll n, ll p, ll mod) {
	if (p == 0 || n == 1) {
		return 1;
	}
	else {
		ll ret = pow2(n * n % mod, p / 2, mod);
		if (p % 2 == 1) {
			ret *= n;
		}
		return ret % mod;
	}
}


#define rep(i,s,e) for(ll i=s;i<e;i++)
#define repeq(i,s,e) for(ll i=s;i<=e;i++)

int main() {
	ll N, M, Q;
	cin >> N >> M >> Q;

	vector<vll> t(N + 2, vll(N + 1, 0));
	rep(i, 0, M) {
		ll l, r;
		cin >> l >> r;
		t[l][r]++;
	}
	for (ll i = N; i >= 0; i--) {
		repeq(j, 1, N) {
			t[i][j] += t[i][j - 1];
		}
		repeq(j, 1, N) {
			t[i][j] += t[i + 1][j];
		}
	}

	rep(i, 0, Q) {
		ll p, q;
		cin >> p >> q;
		cout << t[p][q] << endl;
	}

	return 0;
}