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 Python (3.4.3) (judge verdict: Runtime Error)

#include<iostream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<string>
#include<vector>
#include<stack>
#include<iomanip>
#include<map>
#define rep(i,n) for(int i=0;i<((int)n);i++)
#define drep(i,n) for(int i=((int)n)-1;i>-1;i--)
#define Rep(i,n0,n) for(int i=n0;i<((int)n);i++)
#define Com(i,j,n) for(int i=0;i<n-1;i++)for(int j=i+1;j<((int)n);j++)
#define W(x) cout<<x<<endl;
#define Debug(x) cout<<"#DEBUG "<<x<<endl;
#define INF (1<<29)
#define INF2 (long long)8*(long long)INF*(long long)INF
#define mod 1000000007

typedef long long ll;

using namespace std;


int min(int x,int y){if(y<x)return y;else return x;}
int max(int x,int y){if(y>x)return y;else return x;}
int fact(int x){int i=1;rep(j,x){i*=(j+1);} return i;}
long long fact(long long x){long long i=1;rep(j,x){i*=(j+1);} return i;}
int gcd(int x, int y){if(x%y==0)return y;else if(y%x==0)return x;else return gcd(x%y,y%x);}
ll gcd(ll x, ll y){if(x%y==0)return y;else if(y%x==0)return x;else return gcd(x%y,y%x);}
void say(bool x){if(x){cout<<"Yes"<<endl;}else {cout<<"No"<<endl;}return;}

int dd[500][500],PP[2][100000];
int main()
{
    int n,m,Q;
    cin >>n >>m >>Q;
    int l,r;
    rep(i,m)
    {
        cin>>l>>r;
        l--;
        r--;
        dd[l][r]++;
    }
    rep(i,n)
    {
        rep(j,n-1)
        {
            dd[i][j+1]+=dd[i][j];
        }
    }
    rep(i,n)
    {
        drep(j,n-1)
        {
            dd[j][i]+=dd[j+1][i];
        }
    }
    int p,q;
    rep(i,Q)
    {
        cin>>p>>q;
        PP[0][i]=--p;
        PP[1][i]=--q;
    }
    rep(i,Q)
    {
        W(dd[PP[0][i]][PP[1][i]])
    }
    return 0;
}