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: Wrong Answer)

/*
Hey there stalker....

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Like what you see?
*/
#include <bits/stdc++.h>
#include <vector>
#include <string>
#include <algorithm>
#include <set>
using namespace std;
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define pb push_back
#define pii         pair<ll ,ll>
#define vi          vector<ll>
#define present(container, element) (container.find(element) != container.end()) /* For set n maps */
#define cpresent(container, element) (find(all(container),element) != container.end()) /* For vectors */
#define all(a)      (a).begin(),(a).end()
#define invect(data,n,commands) for(int i = 0;i<n;i++){int tmp;cin>>tmp;data.pb(tmp);commands}
#define inset(data,n,commands) for(int i = 0;i<n;i++){int tmp;cin>>tmp;data.I(tmp);commands}
#define F           first
#define S           second
#define I           insert
#define sz(x)       (ll)x.size()
#define rep(i,a,b)    for(ll  i=a;i<b;i++)
#define repr(i,a,b) for(ll  i=a;i>b;i--)
#define trav(a, x) 	for(auto& a : x)
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define mp          make_pair
#define lcase(a) (char)tolower(a)
#define ll long long int
#define lld long double
#define endl "\n"
#define PI 3.14159265358979323844
#define MOD 1000000007
#define max(a, b) (a < b ? b : a)
#define min(a, b) ((a > b) ? b : a)
const int N = 1e5 + 5;
const int mod=1e9 + 7;
/*	For Debugging	*/
#define DEBUG 		cerr<<"\n>>>I'm Here<<<\n"<<endl;
#define display(x) trav(a,x) cout<<a<<" ";cout<<endl;
#define what_is(x)  cerr << #x << " is " << x << endl;
/* we choose at least k numbers from (0,N), how
 * many sums possible ?
 * k<=n
 * nCr(n,k) + nCr(n,k+1) + ... nCr(n,n) sets,
 * but _not_ every set has distinct sum.
 * Sets of diff size have diff sums.
 * So, how much sets in one size have same sum?
 *
 * Example, k=2, n=4
 * 1+2=3
 * 1+3=4
 * 1+4=5
 * 2+3=5
 * 2+4=6
 * 3+4=7
 * We can build all sums between smallest and biggest.
 * Smallest is 1+2+...+k
 * Biggest is n+n-1+n-2+n-(k-1)
 * ...
 */
int main()
{
   IOS

    int TESTS;
    cin >> TESTS;
    while (TESTS--)
    {
        ll n,k;
        cin >> n >> k;
        int ans=0;

        for(int i=k;i<=n+1;i++)
        {
          int  mini=(i*(i-1))/2;
           int maxi=n*i-(i*(i-1))/2;
            ans+=maxi-mini+1;
            ans%=mod;

        }

        cout<< ans <<endl;


    }
    return 0;
}