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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

class SegmentTree
  def initialize(size, default:, update: -> (nv, _) { nv }, merge:)
    n = 1
    while n < size
      n *= 2
    end
    @size = n
    @default = default
    @update_proc = update
    @merge_proc = merge
    @data = Array.new(n * 2 - 1, default)
  end

  # 0-origin
  def update(k, a)
    k += @size - 1
    @data[k] = @update_proc.call(a, @data[k])
    while k > 0
      k = (k - 1) / 2
      @data[k] = @merge_proc.call(@data[k * 2 + 1], @data[k * 2 + 2])
    end
    nil
  end

  # [a, b)
  # 0-origin
  def query(a, b)
    execute_query(a, b, 0, 0, @size)
  end

  private

  def execute_query(a, b, k, l, r)
    return @default if r <= a || b <= l
    return @data[k] if a <= l && r <= b
    vl = execute_query(a, b, k * 2 + 1, l, (l + r) / 2)
    vr = execute_query(a, b, k * 2 + 2, (l + r) / 2, r)
    @merge_proc.call(vl, vr)
  end
end

N, M, Q = gets.split.map(&:to_i)
lrs = M.times.map { gets.split.map(&:to_i) }
pqs = Q.times.map { gets.split.map(&:to_i) }
list = []
lrs.each do |(l, r)|
  list.push([l, r, -1])
end
pqs.each.with_index do |(p, q), i|
  list.push([p, q, i])
end
sorted = list.sort_by { |(_, b, c)| [b, c] }
st = SegmentTree.new(
  N,
  default: 0,
  update: -> (a, b) { a + b },
  merge: -> (a, b) { a + b }
)
ans = Array.new(Q)
sorted.each do |(a, b, c)|
  if c == -1
    st.update(a - 1, 1)
  else
    sum = st.query(0, b) - st.query(0, a - 1)
    ans[c] = sum
  end
end
ans.each do |sum|
  puts sum
end