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)

#include <bits/stdc++.h>

// #include <atcoder/all>
using namespace std;
// using namespace atcoder;
typedef long long ll;
/* clang-format off */
#define MOD 1000000007
#define INF 1000000000000000000ll
#define REP(i, n) for (ll i = 0, i##_len = (n); i < i##_len; ++i)
#define ALL(a) (a).begin(), (a).end()
//#define __DEBUG__
#ifdef __DEBUG__
#define CH_P(a) cout <<"check_point("<<#a<<")" << "\n";
#define DEBUG(x) cout<<#x<<":"<<x<<"\n"
#define DEBUGS(v) cout << #v << ":";for(auto x:v){cout<<x<<" ";}cout<<"\n"
#endif
#ifndef __DEBUG__
#define CH_P(a) 
#define DEBUG(x) 
#define DEBUGS(v) 
#endif
/* clang-format on */
// using mint = modint1000000007;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n, k;
  cin >> n >> k;
  ll ans = 0;
  ans = (-(n + 1) * (n + 2) * (2 * n + 3) / 6+MOD) % MOD +
        k * (k - 1) * (2 * k - 1) / 6 % MOD +
        ((n + 1) * ((n + 1) * (n + 2) / 2 - k * (k - 1) / 2)) % MOD + n - k + 2;
  cout << ans%MOD << endl;
  return 0;
}