P19490 [AtCoder] We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.
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题目描述

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-----
3 2

-----Sample Output-----
10

The sum can take 10 values, as follows:
- $(10^{100})+(10^{100}+1)=2\times 10^{100}+1$
- $(10^{100})+(10^{100}+2)=2\times 10^{100}+2$
- $(10^{100})+(10^{100}+3)=(10^{100}+1)+(10^{100}+2)=2\times 10^{100}+3$
- $(10^{100}+1)+(10^{100}+3)=2\times 10^{100}+4$
- $(10^{100}+2)+(10^{100}+3)=2\times 10^{100}+5$
- $(10^{100})+(10^{100}+1)+(10^{100}+2)=3\times 10^{100}+3$
- $(10^{100})+(10^{100}+1)+(10^{100}+3)=3\times 10^{100}+4$
- $(10^{100})+(10^{100}+2)+(10^{100}+3)=3\times 10^{100}+5$
- $(10^{100}+1)+(10^{100}+2)+(10^{100}+3)=3\times 10^{100}+6$
- $(10^{100})+(10^{100}+1)+(10^{100}+2)+(10^{100}+3)=4\times 10^{100}+6$

输入格式

见题面

输出格式

见题面

样例输入
3 2
样例输出
10