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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(non_snake_case)]
#![allow(unused_imports)]
use proconio::input;
use std::cmp::max;
use std::cmp::min;
use std::collections::HashMap;
use std::collections::VecDeque;

// 自力で解こうとしたがうまくいかなかったので答えを見た
// O(NQ+N^2)解法

fn main() {
    input! {
        N: usize,
        M: usize,
        Q: usize,
        tmpLRs: [(usize, usize); M],
        tmpPQs: [(usize, usize); Q],
    }

    // 0-origin
    let LRs: Vec<(usize, usize)> = tmpLRs.iter().map(|&(l, r)| (l - 1, r - 1)).collect();
    let PQs: Vec<(usize, usize)> = tmpPQs.iter().map(|&(p, q)| (p - 1, q - 1)).collect();

    // O(10^5)のループの内側でクエリに回答する
    // つまり、クエリへの回答はO(1)でできる必要がある (← 注: これ嘘です。10^7までは食えるので今回はO(NQ)までいけます)
    // ある「領域」に属する個数を高速で求める → 累積和
    // ここで、領域は幅を持っている → 2次元累積和
    // この問題ではオーバーキルな解法らしい
    // なので、いったんN <=500、つまり N^2 < 10^6を利用した解法を使う

    // x[l][r]を区間[l,r]を走る電車の本数とするとき、
    // acc[l][r+1] := sum $ [ x[l][i] for i in 0..r ]
    let mut acc: Vec<Vec<i64>> = vec![vec![0i64; N + 1]; N];

    // まず目印を付けて、後でaccに直す
    let mut mark: Vec<Vec<i64>> = vec![vec![0i64; N]; N];
    for &(l, r) in LRs.iter() {
        mark[l][r] += 1;
    }

    // accに直す
    // 横一列の累積和でO(N)、それをN行重ねるからO(N^2)
    for l in 0..N {
        for r in 0..N {
            acc[l][r + 1] = acc[l][r] + mark[l][r];
        }
    }

    for &(p, q) in PQs.iter() {
        // ループの中はほぼO(1)じゃないと間に合わない (← 注: これ嘘です。10^7までは食えるので今回はO(NQ)までいけます)
        // 区間、O(1)と来れば累積和でしょう

        let mut ans = 0;
        for l in p..=q {
            ans += acc[l][q + 1] - acc[l][l];
        }
        println!("{}", ans);
    }
}