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 <bits/stdc++.h>
#define REP(i,n) for(int i=0;i<(int)(n);i++)
#define REPS(i,n) for(int i=1;i<=(int)(n);i++)
using namespace std;
using ll = long long;
using ld = long double;

const int MOD = 1000000007;

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

/*
    long long sumOfCombinations =
        (N+N-1) - (0+1) + 1 + // 5
        (N+N-1+N-2) - (0+1+2) + 1 + // 4
        (N+N-1+N-2+N-3) - (0+1+2+3) + 1; // 1
*/
    long long sum[220000];
    for (long long i = 1; i <= N+1; i++) {
        if (i == 1) {
            sum[i] = N + 1;
            continue;
        }
        long long d = N * i - ((i - 1) * i) + 1;
        sum[i] = sum[i-1] + d;
    }

    long long ans = sum[N+1] - sum[K-1];
    ans %= MOD;

    cout << ans << endl;
}