Problem:
Score : 400 points

Problem Statement

We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.

We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).

Constraints

1 \leq N \leq 2\times 10^5

1 \leq K \leq N+1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of possible values of the sum, modulo (10^9+7).

Sample Input 1

3 2

Sample Output 1

10

The sum can take 10 values, as follows:

(10^{100})+(10^{100}+1)=2\times 10^{100}+1

(10^{100})+(10^{100}+2)=2\times 10^{100}+2

(10^{100})+(10^{100}+3)=(10^{100}+1)+(10^{100}+2)=2\times 10^{100}+3

(10^{100}+1)+(10^{100}+3)=2\times 10^{100}+4

(10^{100}+2)+(10^{100}+3)=2\times 10^{100}+5

(10^{100})+(10^{100}+1)+(10^{100}+2)=3\times 10^{100}+3

(10^{100})+(10^{100}+1)+(10^{100}+3)=3\times 10^{100}+4

(10^{100})+(10^{100}+2)+(10^{100}+3)=3\times 10^{100}+5

(10^{100}+1)+(10^{100}+2)+(10^{100}+3)=3\times 10^{100}+6

(10^{100})+(10^{100}+1)+(10^{100}+2)+(10^{100}+3)=4\times 10^{100}+6

Sample Input 2

200000 200001

Sample Output 2

1

We must choose all of the integers, so the sum can take just 1 value.

Sample Input 3

141421 35623

Sample Output 3

220280457

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;

public class Main {

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        int K = scanner.nextInt();
        int[] numbers = new int[N+1];
        for (int i=0; i<=N; i++) {
            numbers[i] = i;
        }

        long sum = 0;

        long min = sum(numbers, K);
        long max = sumReverse(numbers, K);
        sum += max - min + 1;

        for (int i=K; i<=N; i++) {
            min = min + i;
            max = max + N - i;
            sum += max - min + 1;
        }
        System.out.println(sum % 1000000007);
    }
    static long sum(int[] array, int length) {
        long sum = 0;
        for (int i=0; i<length; i++) {
            sum += array[i];
        }
        return sum;
    }
    static long sumReverse(int[] array, int length) {
        long sum = 0;
        for (int i=array.length-1; i > array.length - 1 - length; i--) {
            sum += array[i];
        }
        return sum;
    }
}