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 PyPy3 (2.4.0) (judge verdict: Accepted)

import sys
N, M, Q = map(int, sys.stdin.readline().rstrip().split())

#da[i][j]:(0,0)~(i,j)の長方形の和
def da_generate(h, w, a):
    da = [[0] * w for j in range(h)]
    da[0][0] = a[0][0]
    for i in range(1, w):
        da[0][i] = da[0][i - 1] + a[0][i]
    for i in range(1, h):
        cnt_w = 0
        for j in range(w):
            cnt_w += a[i][j]
            da[i][j] = da[i - 1][j] + cnt_w
    return da

#da_calc(p,q,x,y):(p,q)~(x,y)の長方形の和
def da_calc(p, q, x, y):
    if p > x or q > y:
        return 0
    if p == 0 and q == 0:
        return da[x][y]
    if p == 0:
        return da[x][y] - da[x][q - 1]
    if q == 0:
        return da[x][y] - da[p - 1][y]
    return da[x][y] - da[p - 1][y] - da[x][q - 1] + da[p - 1][q - 1]

a = [[0] * N for _ in range(N)]

for _ in range(M):
    l, r = map(int, sys.stdin.readline().rstrip().split())
    a[l - 1][r - 1] += 1

da = da_generate(N, N, a)

for _ in range(Q):
    p, q = map(int, sys.stdin.readline().rstrip().split())
    p -= 1
    q -= 1
    print(da_calc(p, p, q, q))