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>
using namespace std;
#define rep(i,n) for(int i = 0;i<n;i++)
#define erep(i,n) for(int i = 0;i<=n;i++)
#define rep1(i,n) for(int i = 1;i<n;i++)
#define erep1(i,n) for(int i = 1;i<=n;i++)
typedef long long ll;
#define vint vector<int>
#define vvint vector<vector<int>>
#define vstring vector<string>
#define vdouble vector<double>
#define vll vector<ll>:
#define vbool vector<bool>
#define INF 1101010101010101010
#define MOD 1000000007
#define int long long
using P = pair<int,int>;

signed main(){
  int a;
  int b;
  cin >> a >> b;
  a++;
  int A = 0;
  int B = 0;
  rep(i,b){
    A += i+1;
    B += a-i;
  }
  int ans = 0;
  for(int i = b;i<= a;i++){
    //cout << A << " " << B << " " << B - A << endl;
    ans += B-A+1;
    ans %= MOD;
    A += i+1;
    B += a-i;
  }
  cout << ans << endl;

}