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>
#include <chrono> 
 
using namespace std;
using namespace std::chrono; 
 
// #pragma GCC target ("avx2")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")
// #pragma optimization_level 3
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#define f0r(a, b) for (long long a = 0; a < (b); ++a)
#define f1r(a, b, c) for (long long a = (b); a < (c); ++a)
#define f0rd(a, b) for (long long a = (b); a >= 0; --a)
#define f1rd(a, b, c) for (long long a = (b); a >= (c); --a)
#define ms(arr, v) memset(arr, v, sizeof(arr))
#define pb push_back
#define send {ios_base::sync_with_stdio(false);}
#define help {cin.tie(NULL); cout.tie(NULL);}
#define fix(prec) {cout << setprecision(prec) << fixed;}
#define mp make_pair
#define f first
#define s second
#define presum(p, a, n) {p[0] = a[0]; for (int i = 1; i < (n); i++) p[i] = a[i] + p[i-1];}
#define all(v) v.begin(), v.end()
#define getunique(v) {sort(all(v)); v.erase(unique(all(v)), v.end());}
#define readgraph(list, edges) for (int i = 0; i < edges; i++) {int a, b; cin >> a >> b; a--; b--; list[a].pb(b); list[b].pb(a);}
#define ai(a, n) for (int ele = 0; ele < n; ele++) cin >> a[ele];
#define ain(a, lb, rb) for (int ele = lb; ele <= rb; ele++) cin >> a[ele];
#define ao(a, n) {for (int ele = 0; ele < (n); ele++) { if (ele) cout << " "; cout << a[ele]; } cout << '\n';}
#define aout(a, lb, rb) {for (int ele = (lb); ele <= (rb); ele++) { if (ele > (lb)) cout << " "; cout << a[ele]; } cout << '\n';}
typedef long long ll;
typedef double ld;
typedef long double lld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
 
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v);
template<typename A, typename B> ostream& operator<<(ostream &cout, pair<A, B> const &p) { return cout << "(" << p.f << ", " << p.s << ")"; }
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v) {
  cout << "["; for(int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]";
}
template<typename A, typename B> istream& operator>>(istream& cin, pair<A, B> &p) {
  cin >> p.first;
  return cin >> p.second;
}
 
mt19937 rng(steady_clock::now().time_since_epoch().count());
// mt19937 rng(61378913);
/* usage - just do rng() */
 
void usaco(string filename) {
  // #pragma message("be careful, freopen may be wrong")
	freopen((filename + ".in").c_str(), "r", stdin);
	freopen((filename + ".out").c_str(), "w", stdout);
}
 
const lld pi = 3.14159265358979323846;
const ll mod = 1000000007;
// const ll mod = 998244353;
// ll mod;
 

 
ll n, m, k, q, l, r, x, y, z;
ll a[1000005];
ll b[1000005];
ll c[1000005];
string s, t;
ll ans = 0;



void solve(int tc) {
  cin >> n >> k; ++n;

  ll ans = 1;
  f1r(i, k, n) {
    ll mn = i * (i - 1) / 2;
    ll mx = n * (n - 1) / 2 - (n - i) * (n - i - 1) / 2;
    ans = (ans + mx - mn + 1) % mod;
  }

  cout << ans << '\n';
}
 
int main() {
  send help
 
  // usaco("file");
 
  int tc;
  tc = 1;
  // cin >> tc;
  for (int t = 0; t < tc; t++) solve(t);
}