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: Runtime Error)

#include <iostream>
#include <vector>
#include <math.h>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <string>
#include <cstring>
#include <regex>
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using pii = pair<int,int>;
using pl4 = pair<ll,ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vs = vector<string>;
using vvs = vector<vs>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vpii = vector<pii>;
using vvpii = vector<vpii>;
using vpl4 = vector<pl4>;
using vvpl4 = vector<vpl4>;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pob pop_back()
#define sz size()
#define be begin()
#define en end()
#define asn assign
#define emp empty()
#define ft front()
#define bk back()
#define clr clear()
#define ins insert
#define ers erase
#define res resize

#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define rFOR(i,a,b) for(int i=(b);i>=(a);i--)
#define SORT(a) sort((a).be,(a).en)
#define rSORT(a) sort((a).rbegin(),(a).rend())
#define UNIQUE(a) (a).erase(unique((a).be,(a).en),(a).en)
#define PREVP(a) prev_permutation((a).be,(a).en)
#define NEXTP(a) next_permutation((a).be,(a).en)
#define BINS(a,b) binary_search((a).be,(a).en,(b))
#define LOWB(a,b) (lower_bound((a).be,(a).en,(b))-(a).be)
#define UPB(a,b) (upper_bound((a).be,(a).en,(b))-(a).be)
#define CNT(a,b) count((a).be,(a).en,b)
#define SUM(a) accumulate((a).be,(a).en,0)
#define REV(a) reverse((a).be,(a).en)
#define REGS(a,b) regex_search((a),regex(b))
#define REGM(a,b) regex_match((a),regex(b))
#define yn(a) cout <<((a)?"yes":"no")<<endl;
#define Yn(a) cout <<((a)?"Yes":"No")<<endl;
#define YN(a) cout <<((a)?"YES":"NO")<<endl;
#define say(a) cout <<(a);
#define sal(a) cout <<(a)<<endl;
#define sak cout <<endl;
#define dbg(a) cout <<(#a)<<": "<<(a)<<endl;
#define a2l(a) ((ll)(a-97))
#define A2l(a) ((ll)(a-65))
#define l2a(a) ((char)(a+97))
#define l2A(a) ((char)(a+65))
#define DigN2(a) ((llabs(a)==0)?(1):((ll)(log2(double(llabs(a))))+1))
#define DigN10(a) ((llabs(a)==0)?(1):((ll)(log10(double(llabs(a))))+1))
#define Dig2(a,b) (((a)>>(b))&1)
#define Pow2(a) (1<<(a))
#define llin(a) ll (a);cin >>(a);
#define stin(a) string (a);cin >>(a);
#define rdn(a,b) ((a)/(b))
#define rou(a,b) ((((double(a)/double(b))-((a)/(b)))<0.5)?((a)/(b)):(((a)/(b))+1))
#define rup(a,b) ((((a)%(b))==0)?((a)/(b)):(((a)/(b))+1))
#define min(a,b) ((a<b)?(a):(b))
#define max(a,b) ((a>b)?(a):(b))

#define int ll

const ll MOD = 1e9+7;
const string alp = "abcdefghijklmnopqrstuvwxyz";
const string ALP = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

ll gcd(ll a, ll b){if(b==0)return a;return gcd(b,a%b);}
ll lcm(ll a, ll b){return a/gcd(a,b)*b;}

ll salv(vll v){
    say("{");
    FOR(i,0,v.sz-1){
        say(v[i]);
        if(i!=v.sz-1) say(",");
    }
    sal("}")
}

ll DigS10(ll n){
    ll m=0;
    FOR(i,0,DigN10(n)-1){
        m+=(ll)((llabs(n)%(ll)(pow(10.0,(double)(i+1))))/(ll)(pow(10.0,(double)i)));
    }
    return m;
}

ll isP(ll n){
    if(n<=1) return 0;
    FOR(i,2,(ll)sqrt(n)){
        if(n%i==0) return 0;
    }
    return 1;
}

vll FactM(1,1);
vll FactMI(1,1);

ll PowM(ll a,ll b){
    ll ans=1,x=(a%MOD);
    FOR(i,0,DigN2(b)-1){
        if(Dig2(b,i)==1) ans=(ans*x)%MOD;
        if(i!=(DigN2(b)-1)) x=(x*x)%MOD;
    }
    return ans;
}

void CFactM(ll n){
    FOR(i,FactM.sz,n){
        FactM.pb((FactM[i-1]*(i%MOD))%MOD);
    }
    return;
}

void CFactMI(ll n){
    CFactM(n);
    if(FactMI.sz<(n+1)) FactMI.res(n+1,-1);
    if(FactMI[n]==-1) FactMI[n]=PowM(FactM[n],MOD-2);
    rFOR(i,1,n-1){
        if(FactMI[i]!=-1) break;
        FactMI[i]=((FactMI[i+1]*((i+1)%MOD))%MOD);
    }
    return;
}

ll CombM(ll n,ll k){
    if((n<0)||(k<0)) return 0;
    if(n<k) return 0;
    if(n+1>FactMI.sz) CFactMI(n);
    return ((((FactMI[k]*FactMI[n-k])%MOD)*FactM[n])%MOD);
}

signed main() {
   
  llin(n);
  llin(m);
  llin(q);
  vpl4 l(m);
  vll f(n);
  FOR(i,0,m-1){
      cin>>l[i].fi;
      cin>>l[i].se;
      f[l[i].fi]++;
  }
  SORT(l);
  FOR(i,1,n-1){
      f[i]=f[i-1]+f[i];
  }
  ll p;
  ll u;
  ll k;
  FOR(i,0,u-1){
      k=0;
      cin>>p;
      cin>>u;
      FOR(i,f[p-1],f[u-1]){
          if(l[i].fi<=u) k++;
      }
      sal(k);
  }
  
}