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)

//nani?
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
#include <ext/pb_ds/detail/standard_policies.hpp>
#define pi acos(-1);
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define f(i,a,b) for( int i = a; i < b ; i++ )
#define sor(a) sort( a.begin(), a.end() )
#define rsor(a) sort( a.rbegin(), a.rend() )
#define fastio ios_base::sync_with_stdio(0);cin.tie(0)
//#define inf 1LL<<62
typedef long long ll;
typedef double ld;
using namespace std;
using namespace __gnu_pbds;
const int inf=1e7;
const ll MOD=1e9+7;
const ll mod=998244353;
// ac cmtr;
const int MAX_N = 2e5+5;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;

#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << endl;}
 
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1<<" | ";
  __f(comma+1, args...);}
#else
#define trace(...)
#endif
//---------------------------

ll bp(ll b,ll e){
	if(e==0) return 1LL;
  	ll aux=bp(b,e/2);
  	if(e%2==1){
    	return (((b*aux)%MOD)*aux)%MOD;
    }
  	else{
      return (aux*aux)%MOD;
    }
}
int main(){
    fastio;
    ll n,k;
  	cin>>n>>k;
  	ll ans=0;
  	ll val = n*(n+1)/2;
  	for(ll i=k;i<=n+1;i++){
      	ll val1 = (n-i)*(n-i+1)/2;
      	ll val2 = (i-1)*i/2;
      	//trace(i,val,val1,val2);
    	ans += ((val-val1)-val2+1);
      	ans%=MOD;
    }
  	cout<<ans<<endl;
  
    
    
}