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: Runtime Error)

#include "bits/stdc++.h"
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define all(s) s.begin(), s.end()
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
typedef vector<string> vs;
const int MX = 100005, INF = 1001001001;
const ll LINF = 1e18;
const double eps = 1e-10;
const int DIV = 2e9+7;

int main() {
    int n, m, q;
    cin >> n >> m >> q;
    vvi roads(m);
    rep(i, m) {
        int from, to;
        cin >> from >> to;
        from--; to--;
        roads[from].pb(to);
    }
    vvi end_nums(n);
    rep(from_ptr, n) {
        int cnt = 0;
        rep(end_ptr, n) {
            for(auto t: roads[from_ptr]) {
                if (t == end_ptr) cnt++;
            }
            end_nums[from_ptr].pb(cnt);
        }
    }

    rep(i, q) {
        int qfrom, qto;
        cin >> qfrom >> qto;
        qfrom--; qto--;
        int ans = 0;
        for(int start = qfrom; start <= qto; start++) {
            ans += end_nums[start][qto];
        }
        // cout << qfrom << " -> " << qto << " : " << ans << endl;
        cout << ans << endl;
    }
}