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: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n) for (decltype(n) i = 0; i < n; i++)
#define REPvec(itr, mp) for (auto itr = mp.begin(); itr != mp.end(); itr++)
#define readvec(vec)    \
  for (auto& e : vec) { \
    cin >> e;           \
  }
#define all(x) x.begin(), x.end()
#define MOD 1000000007  // 1e9+7
using namespace std;
typedef long long int ll;
typedef vector<int> vecint;
typedef vector<ll> vecll;
typedef vector<double> vecd;
typedef vector<string> vecstr;
typedef vector<char> vecchar;
typedef vector<bool> vecbool;
typedef vector<vector<bool>> vec2bool;
typedef vector<vector<int>> vec2int;
typedef vector<vector<double>> vec2d;
////////////////////////////////////////////////
void init() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}
////////////////////////////////////////////////
int main() {
  init();
  int n, m, q;
  cin >> n >> m >> q;
  vector<vector<int>> g(n + 1);
  vecint l(m), r(m), x(q), y(q);
  vecint ans(q, 0);
  REP(i, m) cin >> l[i] >> r[i];
  REP(i, q) cin >> x[i] >> y[i];
  REP(i, m) { g[l[i]].push_back(r[i]); }
  // cout << endl;

  // REP(i, n) {
  //   if (g[i + 1].empty()) continue;
  //   cout << i + 1 << ": ";
  //   for (auto& e : g[i + 1]) cout << e << " ";
  //   cout << endl;
  // }

  REP(i, q) {
    int num = 0;
    int a = x[i];
    int b = y[i];
    // cout << "a = " << a << ", b = " << b << endl;
    for (int j = a; j <= b; j++) {
      if (g[j].size() == 0) continue;
      if (g[j][0] > b) continue;
      if (g[j].back() <= b) {
        num += g[j].size();
        continue;
      }
      auto itr = upper_bound(all(g[j]), b);
      num += distance(g[j].begin(), itr);
      // if (i == 1) cout << distance(g[j].begin(), itr) << endl;
    }
    ans[i] = num;
  }
  // cout << endl;

  for (auto& e : ans) cout << e << endl;
}