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)

//Created by Kira

#include<bits/stdc++.h>
using namespace std;

#define IOS cin.sync_with_stdio(0);cin.tie(0); cout.tie(0);
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
#define pii pair<int,int> 
#define pb push_back
#define vi vector<int> 
using ll = long long;

const int mod = 1e9 + 7;

ll pwr(ll a, ll b);
	



int main(){
   IOS

	int n, k; cin>>n>>k;

	vector<ll> v;
	v.pb(0); v.pb(0);
	for(int i = 1; i <= n; i++){
		v.pb(v.back() + i);
	}

	ll ans = 0;
	for(int r = k; r <= n+1; r++){
		ll lo = v[r] - v[0];
		ll hi = v[n + 1] - v[n + 1 - r];
		//cout<<lo<<" "<<hi<<"\n";
		ans += (hi - lo + 1 + mod)%mod;
		if(ans >= mod)ans -= mod;
	}

	cout<<ans<<"\n";







}

ll pwr(ll a, ll b){
   a %= mod;	//Remove mod if not required
   ll res = 1;
   while(b > 0){
	if(b&1)
		res = res * a % mod;
		a = a * a % mod;
	b >>= 1;
	}

	return res;
}