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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

class Main{

	static int N;
	static long M;

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);			//文字の入力

		int N = sc.nextInt();
		int M = sc.nextInt();
		int Q = sc.nextInt();
		int[] L  = new int[M];
		int[] R  = new int[M];

		int[][] memo = new int[N+1][N+1];

		for(int i= 0;i < M;i++){
			L[i] = sc.nextInt();
			R[i] = sc.nextInt();
			memo[L[i]][R[i]]++; 
		}

		int[][] dp = new int[N+1][N+1];

		for(int i = 1;i <= N;i++){
			for(int j = 1;j <= N;j++){
				dp[i][j] = dp[i][j-1] + memo[i][j];
			}
		}


		for(int i= 1;i <= Q;i++){
		int	p = sc.nextInt();
		int	q = sc.nextInt();
int sum = 0;
			for(int j = 1;j <= q;j++){
			 sum += (dp[j][q] - dp[j][p-1]);
			}

			System.out.println(sum);
		}




	}
	static int upperbond(int k, int[] data){		//kより小さい最大の値をdataの中から探し、返す二分探索
		int min = -1;
		int max = data.length;

		while(max-min > 1){
			int mid = (max+min)/2;
			if(data[mid] > k){
				max = mid;
			}else{
				min = mid;
			}
		}
		return max;
	}

	static int lowerbond(int k, int[] data){		//kより小さい最小の値をdataの中から探し、返す二分探索
		int min = -1;
		int max = data.length;

		while(max-min > 1){
			int mid = (max+min)/2;
			if(data[mid] >= k){
				max = mid;
			}else{
				min = mid;
			}
		}
		return min+1;
	}





}