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)

#define ONLINE_JUDGE
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> pl;
typedef vector<pl> vp;
const ll INF=1001001001;
const ll LINF=1001001001001001001;
const ll D4[]={0,1,0,-1,0};
const ll D8[]={0,1,1,0,-1,-1,1,-1,0};
#define _overload3(_1,_2,_3,name,...) name
#define _rep2(i,n) for(ll i=0;i<(n);++i)
#define _rep3(i,a,b) for(ll i=(a);i<(b);++i)
#define rep(...) _overload3(__VA_ARGS__,_rep3,_rep2)(__VA_ARGS__)
#define _repe2(i,n) for(ll i=0;i<=(n);++i)
#define _repe3(i,a,b) for(ll i=(a);i<=(b);++i)
#define repe(...) _overload3(__VA_ARGS__,_repe3,_repe2)(__VA_ARGS__)
#define _rrep2(i,n) for(ll i=(n)-1;i>=0;i--)
#define _rrep3(i,a,b) for(ll i=(b)-1;i>=(a);i--)
#define rrep(...) _overload3(__VA_ARGS__,_rrep3,_rrep2)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
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(b<a){a=b;return 1;}return 0;}
template<class T>
using pq=priority_queue<T,vector<T>,greater<T>>;
const ll mod = 1000000007;

ll N;
ll K;

void solve();

int main(){
    cin.tie(0);
    ios::sync_with_stdio(0);
    cin >> N;
    cin >> K;
    solve();
}

// 1 <= N <= 2× 10^5
// 1 <= K <= N+1
// 入力は全て整数

// ll N, ll K
// 和としてあり得るものの個数を \bmod (10^9+7) で出力せよ。
void solve(){
    ll ma=0,mi=0,ans=0;
    rep(i,0,N+1){
        ma+=N-i;
        mi+=i;
        if(i>=K-1)(ans+=ma-mi+1)%=1000000007;
    }
    cout<<ans<<endl;
}