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 <bits/stdc++.h>
#define rep(i,n) for (ll i = 0; i < n; ++i)
#define sz(x) int(x.size())
#define ALL(c) (c).begin(), (c).end()
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v, x) (std::find(v.begin(), v.end(), x) != v.end())
#define FORV(i, v) for (auto i = v.begin(); i != v.end(); i++)
using namespace std;

using ll = long long;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
const int INF = 1001001001;
const long long INFLL = LLONG_MAX / 3;
const double eps = (1e-9);

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n, m, q;
  cin >> n >> m >> q;
  vector<int> lr(m);
  rep(i, m) {
    int l, r; cin >> l >> r;
    lr[i] = l * 1000 + r;
  }

  sort(ALL(lr));

  rep(i, q) {
    int p, q;
    cin >> p >> q;
    auto itr1 = lower_bound(ALL(lr), p * 1000);
    auto itr2 = lower_bound(ALL(lr), (q+1) * 1000);
    int ans = 0;
    while (itr1 != itr2) {
      if (*itr1 % 1000 <= q) ans++;
      itr1++;
    }
    cout << ans << endl;
  }

  return 0;
}