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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>  /*入出力*/
#include <stdlib.h> /*標準ライブラリ*/
#include <string.h> /*文字列*/
#include <math.h>   /*計算*/
#include <assert.h> /*確認*/
#include <ctype.h>  /*文字操作*/
#include <stdbool.h>/*bool型*/
#include <limits.h>/*限界*/
//---------------------------------------------------
#define rep(i, n) for(int  i = 0; i < (n); i++)
#define max(a,b) ((a>b)?a:b)
#define ABS(a) (((a)>(0))?a:-(a))
#define LL long long
#define MOD 1000000007//10^9+7(素数)
//---------------------------------------------------
void swap(int *a, int *b)
{
  int temp;
  temp = *a;*a = *b;*b = temp;
}
//---------------------------------------------------

int main(void)
{
  LL n;
  LL m;
  LL ans = 0;
  scanf("%lld %lld",&n,&m);
  for (LL i = m; i <= n+1; i++) {
    ans += ((2*n - i + 1) - (i-1))*i/2 + 1;
    ans %= MOD;
  }
  printf("%lld\n",ans );

  return 0;
}