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 pb push_back
#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,n,s) for(int i=(s);i<(n);i++)
#define rrep(i,n) for(int i=(n-1);i>=0;i--)
#define rreps(i,n,s) for(int i=s;i>=n;i--)
#define MOD 1000000007
using ll = long long;
using namespace std;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

int main(){
cin.tie(0);
ios::sync_with_stdio(false);
ll n, k;
cin >> n >> k;
vector<ll> cum(200005,0);
reps(i,200005,1){
    cum[i] =cum[i-1]+ i-1;
}

//選ぶ数字の数毎に組合せは重複しない
ll ans = 0;
reps(i,n+2,k){
    ll _min = cum[i];
    ll _max = cum[n+1]-cum[n+1-i];
    ans = (ans + _max - _min + 1)%MOD;
}

cout << ans << endl;

//system("pause");
return 0;
}