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)

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>

#define INF 1000000021
#define MOD 1000000007
#define pb push_back
#define sqr(a) (a)*(a)
#define M(a, b) make_pair(a,b)
#define T(a, b, c) make_pair(a, make_pair(b, c))
#define F first
#define S second
#define all(x) (x.begin(), x.end())
#define deb(x) cerr << #x << " = " << x << '\n'
#define N 222222

using namespace std;
//using namespace __gnu_pbds;

typedef long double ld;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;

const ld pi = 2 * acos(0.0);
template<class T> bool umin(T& a, T b){if(a>b){a=b;return 1;}return 0;}
template<class T> bool umax(T& a, T b){if(a<b){a=b;return 1;}return 0;}
template<class T, class TT> bool pal(T a, TT n){int k=0;for(int i=0;i<=n/2;i++){if(a[i]!=a[n-i-1]){k=1;break;}}return k?0:1;}

//int month[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

int a[501][501];
int main()
{
    int n, m, q;
    cin >> n >> m >> q;
    for(int i = 1; i <= m; i ++)
    {
        int l, r;
        cin >> l >> r;
        a[l][r] ++;
    }
    for(int i = 1; i <= n; i ++)
    for(int j = 1; j <= n; j ++)
    	a[i][j] += a[i][j - 1];
    for(int i = 1; i <= q; i ++)
    {
        int l, r;
        cin >> l >> r;
        int s = 0, mx = 0;
        for(int j = l; j <= r; j ++)
        {
        	s += a[j][r] - a[j][l - 1];
        }
//        deb(mx);
//        deb(s);?
        cout << s << '\n';
    }










    getchar();
    getchar();
    return 0;
    //ios::sync_with_stdio(false);
    //cin.tie(0);
}