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)

/******************************************************************************
 
рассвет. 
 
*******************************************************************************/
#include <bits/stdc++.h>
typedef long long int lli;
typedef unsigned long long int ulli;
typedef long double ld;
#define rapgame ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb push_back
#define eb emplace_back
#define vi vector <int>
#define vlli vector <lli>
#define vulli vector <ulli>
#define vchar vector <char>
#define vstr vector <string>
#define vb vector <bool>
#define fst first
#define sec second
#define srt(a) sort(a.begin(),a.end())
#define rev(a) reverse(a.begin(),a.end())
#define vpii vector <pair <int,int> >
#define pii pair <int,int> 
#define vplli vector <pair <lli,lli> >
#define vpulli vector <pair <ulli,ulli> >
#define vvi vector <vector <int> >
#define vvlli vector <vector <lli> >
#define vvpii vector <vector <pair <int,int>> >
#define fr(i,n) for (int i = 0; i < (n); ++i)
#define frbig(i,n) for (lli i = 0; i < (n); ++i)
#define fra(i,a,n) for (lli i = a; i < (n); ++i)
#define run(c,s) for(auto c:s) 
#define read(n) int n; cin >> n;
#define readchar(n) char n; cin >> n;
#define readstr(n) string n; cin >> n;
#define readarr(a,n) fr(i,n) cin >> a[i];
#define read2(x,y) int x,y; cin >> x >> y;
#define read2ll(x,y) long long int x,y; cin >> x >> y;
#define readll(n) lli n; cin >> n;
#define watch(x) cout << (#x) << " is " << (x) << endl
#define print_array(x) fr(i,x.size()) cout << x[i] << " ";
using namespace std;
 
const lli MOD=1e9+7;
const lli MAXN=1e6+1;
const ld PI=acos(-1);
 
int main(){
    rapgame
    read2ll(n,k);
    lli ans=0;
    fra(i,k,n+2){
        ans+=n*i-(i-1)*i+1;
        ans%=MOD;
    }
    cout << ans;
    return 0;
}