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>
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL);
#define ll long long
#define ull unsigned long long
#define FI(i,a,n) for(int i=a; i<=n; i++)
#define RFI(i,n,a) for(int i=n; i>=a; i--)
#define FLL(i,a,n) for(ll i=a; i<=n; i++)
#define RFLL(i,n,a) for(ll i=n; i>=a; i--)
#define vi vector<int>
#define vll vector<ll>
#define all(v) v.begin(),v.end()
#define pll pair<ll,ll>
#define pi pair<int, int>
#define GCD(a,b) __gcd(a,b)
#define MP make_pair
#define PB push_back
#define PF push_front
#define LB lower_bound
#define UB upper_bound
#define debug(x) cout << x << endl
#define MOD 1000000007
#define INF LLONG_MAX
#define PI 3.14159265359
using namespace std;


ll power(ll x, ll n)
{
  ll ans=1;
  while(n)
  {
    if(n&1) ans = (ans*x)%MOD;
    n>>=1;
    x = (x*x)%MOD;
  }
  return ans;
}

ll sum(ll n)
{
  return (n*(n+1)/2)%MOD;
}

int main()
{
  fastio
  ll n, k;
  ll ans=0;
  cin >> n >> k;
  for(ll i=k; i<=n+1; i++)
  {
    ll temp = sum(n) - sum(n-i) - sum(i-1) + 1;
    if(temp < 0 ) temp += MOD;
    temp%=MOD;
    /* cout << temp << "\n"; */  
    ans = (ans + temp)%MOD;
  }
  cout << ans;
  return 0;
}