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: Wrong Answer)

def s()
  gets().chomp
end
def i()
  gets.to_i
end
def li()
  gets.split.map(&:to_i)
end
def ili()
  n = gets.to_i
  as = gets.split.map(&:to_i)
  [n, as]
end
class Array
  def sum
    total = 0
    self.each do |item|
      total += item
    end
    total
  end
end

def eratosthenes(size)
  prime_list = []
  is_prime = Array.new(size, true)
  is_prime[0] = false
  is_prime[1] = false

  2.upto(size-1){|i|
    if is_prime[i]
      prime_list << i
      (2).upto((size.to_f/i).ceil-1){|j|
        is_prime[j*i] = false
      }
    end
  }
  [prime_list, is_prime]
end

def dijkstra(s, e, cost)
  size = cost.length
  if s==e
    return 0
  end
  dp = {}
  dp[s] = 0
  undefined_node = {}
  size.times{|i|
    if i != s
      undefined_node[i] = cost[s][i]
    end
  }
  while undefined_node.length != 0
    min = undefined_node.min_by{|_, v| v}
    stock = {}
    temp = undefined_node.clone
    temp.each{|k, v|
      if min[1] == v
        undefined_node.delete(k)
        if k == e
          return v
        end
        dp[k] = v
        stock[k] = v
      end
    }

    stock.each{|k, v|
      undefined_node.each{|k2, v2|
        undefined_node[k2] = v + cost[k][k2] < v2 ?  v + cost[k][k2] : v2
      }
    }
  end
end

def get_bits(len)
  res =  []
  (2**len).times{|i|
    res << i.to_s(2).rjust(len, '0')
  }
  res
end

n, m, q = li()
ruiseki = Array.new(501, 0)
501.times{|i|
  ruiseki[i] = Array.new(501, 0)

}


lrs = []
m.times{
  temp = li()
  lrs << temp
  ruiseki[temp[0]][temp[1]] += 1
}

500.times{|i|
  500.times{|j|
    ruiseki[500-j-1][i] += ruiseki[500-j][i]
  }
}
500.times{|i|
  500.times{|j|
    ruiseki[i][j+1] += ruiseki[i][j]
  }
}
qs = []
q.times{
  qs << li()
}

qs.each{|l, r|
  puts ruiseki[l][r]

}