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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define int long long
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define DEC(i, a, b) for (int i = (a); i > (b); --i)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define pb push_back
#define ALL(obj) (obj).begin(), (obj).end()
#define debug(x) cerr << #x << ": " << x << '\n'
using namespace std;
typedef long long ll;
const int INF = (int)1e9;
const int MOD = (int)1e9 + 7;
const ll LINF = (int)1e18;
const double EPS = 1e-9;

const int MAX_N = 1<<17;

// セグメント木を持つグローバル配列
int n, dat[2 * MAX_N - 1];

// 初期化
void init(int n_) {
  // 簡単のため要素数を2のべきに
  n = 1;
  while (n < n_) n*= 2;

  // すべての要素を 0 に
  REP(i,2*n-1) dat[i] = 0;
}

// k番目の値(0-indexed)をaに変更
void update(int k, int a) {
  // 葉の節点
  k += n - 1;
  dat[k] = a;
  // 登りながら更新
  while (k > 0) {
    k = (k - 1) / 2;
    dat[k] = dat[k * 1 + 1] + dat[k * 2 + 2];
  }
}

// [a,b) を完全に含む区間の数を求める
// 後ろのほうの引数は、計算の簡単のための引数
// kは節点の番号、l,rはその節点が[l,r)に対応づいていることを表す。
// したがって外からはquery(a,b,0,0,n)として呼ぶ。
int query(int a, int b, int k, int l, int r) {
  // [a,b) と [l,r) が交差しなければ 0
  if (r <= a || b <= l) return 0;

  // [a,b) が [l,r) を完全に含んでいれば、この節点の値
  if (a <= l && r <= b) return dat[k];
  else {
    // そうでなければ2つの子を再帰的に調べる
    return query(a, b, k * 2 + 1, l, (l + r) / 2) 
      + query(a, b, k * 2 + 2, (l + r) / 2, r);
  }
}

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

  int N,M,Q;
  cin>>N>>M>>Q;
  vector<int> L(M),R(M);
  REP(i,M) {
    cin>>L[i]>>R[i];
  }
  vector<int> p(Q),q(Q);
  REP(i,Q) {
    cin>>p[i]>>q[i];
  }

  REP(i,Q) {
    int ans = 0;
    REP(j,M) {
      if (p[i] <= L[j] && R[j] <= q[i]) ++ans;
    }
    cout << ans << endl;
  }

  return 0;
}