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>
using namespace std;

typedef long long ll;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<vector<ll>>> vvvll;
typedef vector<bool> vb;
typedef vector<vector<bool>> vvb;
typedef vector<vector<vector<bool>>> vvvb;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef vector<vpll> vvpll;
typedef vector<double> vd;
typedef vector<vd> vdd;

#define FOR(i,x,y) for(ll i=(ll)x; i<(ll)y; ++i)
#define REP(i,y) FOR(i, 0, y)
#define RFOR(i,x,y) for(ll i=(ll)x; i>=(ll)y; --i)
#define RREP(i,x) RFOR(i, x, 0)
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define debug_print(x...) cerr << "line " << __LINE__ << " : "; debug_print_in(x);

template <typename First>
void debug_print_in(First first){
  cerr << first << endl;
  return;
}

template <typename First, typename... Rest>
void debug_print_in(First first, Rest... rest){
  cerr << first << " ";
  debug_print_in(rest...);
  return;
}

void IN(void){
  return;
}

template <typename First, typename... Rest>
void IN(First& first, Rest&... rest){
  cin >> first;
  IN(rest...);
  return;
}

template <typename First>
void OUT(First first){
  cout << first << endl;
  return;
}

template <typename First, typename... Rest>
void OUT(First first, Rest... rest){
  cout << first << " ";
  OUT(rest...);
  return;
}

template<class t, class u> void chmax(t&a,u b){if(a<b)a=b;};
template<class t, class u> void chmin(t&a,u b){if(a>b)a=b;};
int popcount(int t){return __builtin_popcount(t);} //GCC
int popcount(ll t){return __builtin_popcountll(t);} //GCC

template <typename T>
void vec_print(vector<T> VEC){
  REP(i, VEC.size()){
    cerr << VEC[i] << " ";
  }
  cerr << endl;
};

template <typename T>
void mat_print(vector<vector<T> > MAT){
  REP(i,MAT.size()){
    REP(j,MAT[i].size()){
      cerr << MAT[i][j] << " ";
    }
    cerr << endl;
  }
};

constexpr int INF = (1<<30);
constexpr ll INFLL = 1LL<<62;
constexpr long double EPS = 1e-12;
constexpr ll MOD = (ll)((1E+9)+7);

int main(){
  cin.tie(0); // cut the cin and cout (default, std::flush is performed after std::cin)
  ios::sync_with_stdio(false); // cut the iostream and stdio (DON'T endl; BUT "\n";)

  ll N, K;
  IN(N,K);

  ll ans = 0;
  FOR(k,K,N+2){
    ll tmp = N*k - k*(k-1) + 1;
    debug_print(k, tmp);
    ans += tmp;
    ans %= MOD;
  }

  OUT(ans);

  return 0;
}