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 Python (3.8.2) (judge verdict: Time Limit Exceeded)

import sys
from itertools import combinations
sys.setrecursionlimit(10 ** 6)  # 再帰関数の再帰の深さを設定
to_index = lambda x: int(x) - 1  # 入力した数字に1を引いたものを返す
print_list_in_2D = lambda x: print(*x, sep="\n")  # リストの要素を改行を挟んで表示する関数

# 入力を整数に変換して受け取る
def input_int():
    return int(input())

def map_int_input():
    return map(int, input())

MII = map_int_input

def MII_split():
    return map(int, input().split())

def MII_to_index():
    return map(to_index, input())

def MII_split_to_index():
    return map(to_index, input().split())

# 入力全てを整数に変換したものの配列を受け取る
def list_int_inputs():
    return list(map(int, input()))

LII = list_int_inputs

def LII_split():
    return list(map(int, input().split()))

# 2次元リスト化
def LII_2D(rows_number):
    return [LII() for _ in range(rows_number)]

def LII_split_2D(rows_number):
    return [LII_split() for _ in range(rows_number)]

N, K = MII_split()

# dp = list(set(sum(n) for n in combinations(range(0, N + 1), K))) # K個選んだときの和の組み合わせ
# for k in range(K+1, N+2):
#     dp = dp + list(set(sum(n) for n in combinations(range(0, N + 1), k)))

# return
# nums = []
count = 0
for num_nums in range(K, N + 2):
    tmp_nums = []
    for n in combinations(range(0, N + 1), num_nums):
        s = sum(n)
        if s not in tmp_nums:
            tmp_nums.append(s)
    count += len(tmp_nums)
    # for i in tmp_nums:
    #     nums.append((num_nums, i))
# print(len(nums))
print(count)