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 PyPy3 (7.3.0) (judge verdict: Accepted)

import sys
readline = sys.stdin.buffer.readline
def even(n): return 1 if n%2==0 else 0

n,k = map(int,readline().split())

mod = 10**9+7

cum = [i for i in range(10**6)]
def cumsum(lst): #元のリストを保持
    res = lst[:]
    for i in range(1,len(res)):
        res[i] += res[i-1]
    return res
cum = cumsum(cum)
def cumsumdif(l,r,lst): #求めたい差のインデックスを入れる
    if l == 0:
        return lst[r]
    else:
        return lst[r] - lst[l-1]

def reydeoro(n):
    return n*(n+1)//2

ans = 0
for i in range(k,n+1):
    ans += cum[n] -cum[n-i] - cum[i-1] + 1 #+1:al自体を含むため
    ans %= mod

print((ans+1)%mod) #k = n+1のとき