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;
typedef long long ll;

const ll maxs = 1e5+5;
const ll lmaxs = 20;
ll mod = 1e9+7;
ll oo = 1e15;

#define IOS ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define I insert
#define endl '\n'

void pre(){

}

ll get(ll x){
    return x*(x-1)/2;
}

void solve(){ 
   ll n , k;
   cin >> n >> k;
   ll ans=0;
   
   for(ll i=k;i<=n+1;i++){
       ll MIN = get(i);
       ll MAX = n*i - get(i);
       ans += MAX - MIN + 1; 
       ans %= mod;
   }

   cout << ans << endl;
}

int main(){
    // #ifndef ONLINE_JUDGE
    // freopen("input00.txt","r",stdin);
    // #endif
    IOS;
    pre();
    ll T = 1;   
    //cin>>T;
    while(T--){  
        solve();
    }
    return 0;
}