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>
#define REP(i,n) for (int i=0;i<(n);i++)
#define FOR(i,s,e) for (int i=s;i<(e);i++)
#define All(v) (v).begin(),(v).end()
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = (int)1e9 + 7;
const int INF = 1010101010;

int ans[510][510];
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N, M, Q;
    cin >> N >> M >> Q;
    REP(i, M){
        int l, r;
        cin >> l >> r;
        ans[l][r]++;
    }
    REP(l, N+1)REP(r, N) ans[l][r+1] += ans[l][r];
    REP(r, N+1)for(int l=N; l>=0; l--) ans[l][r] += ans[l+1][r];

    REP(i, Q){
        int l, r;
        cin >> l >> r;
        cout << ans[l][r] << endl;
    }
    return 0;
}