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>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pll = pair<ll, ll>;
using tll = tuple<ll, ll, ll>;
using pqueue = priority_queue<ll, vector<ll>, greater<ll>>;
template <class T, size_t N1, size_t N2> using array2 = array<array<T, N2>, N1>;
template <class T, size_t N1, size_t N2, size_t N3> using array3 = array<array<array<T, N3>, N2>, N1>;
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,(n)-1,0)
#define FOR(i,a,b) for(int i=(a),i##Len_=(b);i<i##Len_;i++)
#define RFOR(i,a,b) for(int i=(a),i##End_=(b);i>=i##End_;i--)
#define ALL(x) (x).begin(),(x).end()
#define SZ(x) ((ll)x.size())
#define SORT_PAIR_SECOND(x) sort(ALL(x),[](auto& a,auto& b){return a.second<b.second;})
#define SORT_TUPLE_1(x) sort(ALL(x),[](auto& a,auto& b){return get<1>(a)<get<1>(b);})
template <class T> void chmax(T& a, const T& b) { if (b > a) a = b; }
template <class T> void chmin(T& a, const T& b) { if (b < a) a = b; }
const char* LF = "\n";
// ----------------------------------------------------------------
#define DUMP(x) {cout<<#x<<": "<<x<<"\n";}
#define DECLARE_DUMP_CONTAINER(V) template<class T>ostream&operator<<(ostream& o,V<T>&v){cout<<"{ ";for(auto&i:v)cout<<i<<" ";cout<<"}";return o;}
DECLARE_DUMP_CONTAINER(vector);
DECLARE_DUMP_CONTAINER(deque);

template<class T, size_t N> ostream& operator << (ostream& os, array<T, N>& v) {
    cout << "{ "; for (auto& i : v) cout << i << " " ; cout << "}";
    return os;
}

template <class T, class U> ostream& operator << (ostream& os, pair<T, U>& p) {
    cout << "(" << p.first << " " << p.second << ")";
    return os;
}

template <class T, class U, class V> ostream& operator << (ostream& os, tuple<T, U, V>& t) {
    cout << "(" << get<0>(t) << " " << get<1>(t) << " " << get<2>(t) << ")";
    return os;
}
// ----------------------------------------------------------------

namespace{struct _{_(){
    cin.tie(0);
    ios::sync_with_stdio(false);
}}a;}

// ----------------------------------------------------------------

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

    //array2<ll, N+1, N+1> S = {0};
    vector<vector<ll>> S(N+1, vector<ll>(N+1, 0));
    vector<vector<ll>> T(N+1, vector<ll>(N+1, 0));

    REP(i, M) {
        int l, r;
        cin >> l >> r;
        S[l][r]++;
    }

    REP(i, N+1) {
        ll sum = 0LL;
        REP(j, N+1) {
            sum += S[i][j];
            T[i][j] = sum;
        }
    }

    REP(i, Q) {
        int l, r;
        cin >> l >> r;

        ll ans = 0LL;
        FOR(j, l, r+1) {
            ans += T[j][r] - T[j][l-1];
        }
        cout << ans << LF;
    }


    return 0;
}