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)

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//
#include <bits/stdc++.h>
#define int long long
#define ll long long 
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define erep(i,a,b) for(int i=a;i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>(b);--i)
#define eper(i,a,b) for(int i=(a);i>=b;--i)
#define fore(i, x, a) for(auto &&x:a) 
#define ITR(i,b,e) for(auto i=(b);i!=(e);++i)
#define IN(i,a,b) (a)<=(i) && (i)<(b)
#define pb push_back
#define mp make_pair
#define ALL(x) begin(x),end(x)
#define F first
#define S second
#define debug(x)  cout << #x << ": " << (x) << '\n';
const long long INF=1001001001001001001;
const int MOD=(int)1e9 + 7;
const double EPS=1e-9;
using namespace std;
using Pii = pair<int,int>;
using vii = vector<int>;
template<class T>using  PS_queue = priority_queue<T, vector<T>, greater<T> >;
template<class T>using vv = vector<T>;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
template<typename A, size_t N, typename T> void Fill(A (&array)[N], const T &val){fill( (T*)array, (T*)(array+N), val );}
template<class T> ostream& operator<<(ostream &os,const vector<T> &v) {
    ITR(i,begin(v),end(v))os<<*i<<(i==end(v)-1?"":" ");return os;}
template<class T> istream& operator>>(istream &is,vector<T> &v) {
    ITR(i,begin(v),end(v)) is>>*i;return is;}
template<class T,class U> istream& operator>>(istream &is, pair<T,U> &p) {
    is>>p.first>>p.second;return is;}
template<class T>T gcd(T a, T b){ return b ? gcd(b, a % b) : a; }
template<class T>T lcm(T a, T b){ return a / gcd(a, b) * b; }
struct edge { int from, to, cost; };
int dy[]={0, 1, -1, 0}; int dx[]={1, 0, 0, -1};
// cout << fixed;
// cout << setprecision(10) << val;

int n, m, Q, imos[510][510];
signed main() {
 cin.tie(0);
 ios::sync_with_stdio(false);
  cin >> n >> m >> Q;
  rep(i, 0, m) {
    int l, r;
    cin >> l >> r;
    imos[l][r]++;
  }
  erep(i, 1, n) erep(j, 1, n) {
    imos[i][j] += imos[i][j-1];
  }
  erep(i, 1, n) erep(j, 1, n) {
    imos[i][j] += imos[i-1][j];
  }
  rep(LOOP, 0, Q) {
    int p, q;
    cin >> p >> q;
    cout << imos[q][q] - imos[p-1][q] - imos[q][p-1] + imos[p-1][p-1] << endl;
  }
  return 0;
}