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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

// Shortening code
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pi;
typedef tuple<int, int, int> ti;

// Macros
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define MT make_tuple
#define REP(i, a, b) for (auto i = a; i < b; i++)
#define RREP(i, a, b) for (auto i = a; i >= b; i--)
const ll MOD = 1e9 + 7;
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    // Your code
    ll n, k;
    cin >> n >> k;
    ll cnt = 0;
    REP(i, k, n + 2)
    {
        int max_num = n * i - ((i - 1) * i) / 2;
        int min_num = i * (i - 1) / 2;
        cnt += (max_num % MOD - min_num % MOD + 1)%MOD;
    }
    cout << cnt % MOD << endl;
    return 0;
}