#ABC163D. [ABC163D] Sum of Large Numbers

[ABC163D] Sum of Large Numbers

Score : 400400 points

Problem Statement

We have N+1N+1 integers: 1010010^{100}, 10100+110^{100}+1, ..., 10100+N10^{100}+N.

We will choose KK or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (109+7)(10^9+7).

Constraints

  • 1N2×1051 \leq N \leq 2\times 10^5
  • 1KN+11 \leq K \leq N+1
  • All values in input are integers.

Input

Input is given from Standard Input in the following format:

NN KK

Output

Print the number of possible values of the sum, modulo (109+7)(10^9+7).

3 2
10

The sum can take 1010 values, as follows:

  • (10100)+(10100+1)=2×10100+1(10^{100})+(10^{100}+1)=2\times 10^{100}+1
  • (10100)+(10100+2)=2×10100+2(10^{100})+(10^{100}+2)=2\times 10^{100}+2
  • $(10^{100})+(10^{100}+3)=(10^{100}+1)+(10^{100}+2)=2\times 10^{100}+3$
  • (10100+1)+(10100+3)=2×10100+4(10^{100}+1)+(10^{100}+3)=2\times 10^{100}+4
  • (10100+2)+(10100+3)=2×10100+5(10^{100}+2)+(10^{100}+3)=2\times 10^{100}+5
  • $(10^{100})+(10^{100}+1)+(10^{100}+2)=3\times 10^{100}+3$
  • $(10^{100})+(10^{100}+1)+(10^{100}+3)=3\times 10^{100}+4$
  • $(10^{100})+(10^{100}+2)+(10^{100}+3)=3\times 10^{100}+5$
  • $(10^{100}+1)+(10^{100}+2)+(10^{100}+3)=3\times 10^{100}+6$
  • $(10^{100})+(10^{100}+1)+(10^{100}+2)+(10^{100}+3)=4\times 10^{100}+6$
200000 200001
1

We must choose all of the integers, so the sum can take just 11 value.

141421 35623
220280457