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: Accepted)

#include "bits/stdc++.h"
#pragma GCC optimize("Ofast")

// Begin {{{
using namespace std;

#ifndef DEBUG
#define dump(...)
#endif

template <class A, class B> inline bool chmax(A &a, const B &b) { return b > a && (a = b, true); }
template <class A, class B> inline bool chmin(A &a, const B &b) { return b < a && (a = b, true); }

template <class T>
inline vector<T> make_v(const T &initvalue, intmax_t sz) {
    return vector<T>(sz, initvalue);
}

template <class T, class... Args>
inline auto make_v(const T &initvalue, intmax_t sz, Args... args) {
    return vector<decltype(make_v<T>(initvalue, args...))>(sz, make_v<T>(initvalue, args...));
}

constexpr int INF = 0x3f3f3f3f;
constexpr int64_t LINF = 0x3f3f3f3f3f3f3f3fLL;
// }}} End

// CumulativeSum2D {{{
class CumulativeSum2D {
    vector<vector<intmax_t>> cum;

public:
    CumulativeSum2D() {}
    CumulativeSum2D(const size_t h, const size_t w) : cum(h + 1, vector<intmax_t>(w + 1)) {}

    inline void build() {
        const size_t h = cum.size(), w = cum[0].size();
        for (size_t y = 1; y < h; ++y) {
            for (size_t x = 1; x < w; ++x) {
                cum[y][x] += cum[y][x - 1] + cum[y - 1][x] - cum[y - 1][x - 1];
            }
        }
    }

    inline intmax_t sum(size_t sy, size_t sx, size_t gy, size_t gx) const {
        return cum[gy][gx] - cum[sy - 1][gx] - cum[gy][sx - 1] + cum[sy - 1][sx - 1];
    }

    inline intmax_t operator()(size_t y, size_t x) const {
        return cum[y][x];
    }

    inline intmax_t& operator()(size_t y, size_t x) {
        return cum[y][x];
    }

    inline void resize(const size_t h, const size_t w) {
        cum.assign(h + 1, vector<intmax_t>(w + 1));
    }

    inline intmax_t height() const noexcept {
        return cum.size() - 1;
    }

    inline intmax_t width() const noexcept {
        return cum[0].size() - 1;
    }

    friend ostream& operator<<(ostream &out, const CumulativeSum2D &matrix) {
        const size_t h = matrix.cum.size();
        const size_t w = matrix.cum[0].size();
        for (size_t y = 1; y < h; ++y) {
            for (size_t x = 1; x < w; ++x) {
                out << matrix.cum[y][x] << (x == w - 1)[" \n"];
            }
        }
        return out;
    }
};
// }}}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    size_t N, M, Q; cin >> N >> M >> Q;

    CumulativeSum2D cum(N, N);
    for (size_t i = 0; i < M; ++i) {
        size_t L, R; cin >> L >> R;
        cum(L, R)++;
    }

    cum.build();

    while (Q--) {
        size_t p, q; cin >> p >> q;
        cout << cum.sum(p, p, q, q) << "\n";
    }


    return 0;
}