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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#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 srep(i,s,t) for (int i = s; i < t; ++i)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define show(x) cout << #x << " = " << (x) << endl
#define show2(x,y) cout << #x << " = " << (x) << ", " << #y << " = " << (y) << endl
#define show3(x,y,z) cout << #x << " = " << (x) << ", " << #y << " = " << (y) <<  ", " << #z << " = " << (z) << endl
#define showv(v) rep(i,v.size()) printf("%d%c", v[i], i==v.size()-1?'\n':' ')
#define showv2(v) rep(j,v.size()) showv(v[j])
#define showt(t,n) rep(i,n) printf("%d%c", t[i], i==n-1?'\n':' ')
#define showt2(t,r,c) rep(j,r) showt(t[j],c)
#define showvp(p) rep(i,p.size()) printf("%d %d\n", p[i].first, p[i].second)
#define printv(v) rep(i,v.size()) printf("%d\n", v[i])
#define printt(t,n) rep(i,n) printf("%d\n", t[i])
#define incl(v,x) (find(all(v),x)!=v.end())
#define incls(s,c) (s.find(c)!=string::npos)
#define lb(a,x) distance((a).begin(),lower_bound(all(a),(x)))
#define ub(a,x) distance((a).begin(),upper_bound(all(a),(x)))
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define bit(n,k) ((n>>k)&1) // nのk bit目
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define uni(x) x.erase(unique(all(x)),x.end())
#define SP << " " <<
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using vc = vector<char>;
using vvc = vector<vc>;
using vs = vector<string>;
using vb = vector<bool>;
using vvb = vector<vb>;
using P = pair<int, int>;
using T = tuple<int, int, int>;
using vp = vector<P>;
using vt = vector<T>;
const int mod = 1000000007;
const double EPS = 1e-9;
//const long double EPS = 1e-14;
const int INF = (1<<30)-1;
const ll LINF = (1LL<<62)-1;
#define dame { puts("No"); return 0;}
#define yn {puts("Yes");}else{puts("No");}
inline int in() { int x; cin >> x; return x;}
inline ll lin() { ll x; cin >> x; return x;}
inline char chin() { char x; cin >> x; return x;}
inline string stin() { string x; cin >> x; return x;}
inline double din() { double x; cin >> x; return x;}
//template<class T = int> inline T in() { T x; cin >> x; return (x);}
template<typename T>inline ll suma(const vector<T>& a) { ll res(0); for (auto&& x : a) res += x; return res;}
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
char itoa(int n) { return n + '0';}
ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
const int dx[4] = {1, 0,-1, 0};
const int dy[4] = {0, 1, 0,-1};

int main () {
  int n, m, Q;
  cin >> n >> m >> Q;
  vi l(m), r(m);
  vvi x(n,vi(n));
  rep(i,m) {
    cin >> l[i] >> r[i];
    l[i]--, r[i]--;
    x[l[i]][r[i]]++;
  }
  vvi csum(n,vi(n+1));
  rep(i,n) {
    rep(j,n) csum[i][j+1] = csum[i][j] + x[i][j];
  }
  rep(_,Q) {
    int p, q;
    cin >> p >> q;
    p--, q--;
    int ans = 0;
    srep(i,p,q+1) ans += csum[i][q+1] - csum[i][p];
    cout << ans << endl;
  }
  return 0;
}