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 <iostream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <functional>
#include <iterator>
#include <numeric>
#include <cstdint>
#include <cmath>
#include <bitset>
#include <string>
#include <list>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>

#define N_TIMES(i, n)     for ( uint64_t i = 0;     i <  n; ++i )
#define N_TIMES_REV(i, n) for (  int64_t i = n - 1; i >= 0; --i )

using namespace std;

const uint64_t MOD = 1e+9 + 7;

uint64_t sum(uint64_t L, uint64_t R)
{
  uint64_t P = R + L, Q = (R - L) + 1;

  uint64_t S = 0;
  if ((P % 2) == 0) {
    S = ((P / 2) * Q) % MOD;
  } else { // (Q % 2) == 0
    S = (P * (Q / 2)) % MOD;
  }
  return S;
}

uint64_t calc(uint64_t N, uint64_t x)
{
  uint64_t L = sum(0, x - 1);
  uint64_t R = sum(N - (x - 1), N);
  return (((R + MOD) - L) + 1) % MOD;
}

int main()
{
  uint64_t N, K;
  cin >> N >> K;

  uint64_t S = 0;
  for (uint64_t x = K; x <= N + 1; ++x) {
    S = (calc(N, x) + S) % MOD;
  }
  cout << S << endl;

  return 0;
}