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;
#define FOR(var, begin, end) for (int var = (begin); var <= (end); var++)
#define RFOR(var, begin, end) for (int var = (begin); var >= (end); var--)
#define REP(var, length) FOR(var, 0, length - 1)
#define RREP(var, length) RFOR(var, length - 1, 0)
#define EACH(value, var) for (auto value : var)
#define SORT(var) sort(var.begin(), var.end())
#define REVERSE(var) reverse(var.begin(), var.end())
#define RSORT(var) \
  SORT(var);       \
  REVERSE(var)
#define OPTIMIZE_STDIO         \
  ios::sync_with_stdio(false); \
  cin.tie(0);                  \
  cout.precision(10);          \
  cout << fixed
#define endl '\n'

const int INF = 1e9;
const ll MOD = 1e9 + 7;
const ll LINF = 1e18;
struct mint {
  ll x;
  mint() : x(0) {
  }
  mint(ll x) : x((x % MOD + MOD) % MOD) {
  }
  mint& operator+=(const mint& r) {
    if ((x += r.x) >= MOD) {
      x -= MOD;
    }
    return *this;
  }
  mint& operator-=(const mint& r) {
    if ((x -= r.x) < 0) {
      x += MOD;
    }
    return *this;
  }
  mint& operator*=(const mint& r) {
    if ((x *= r.x) > MOD) {
      x %= MOD;
    }
    return *this;
  }
  mint& operator++() {
    if ((++x) >= MOD) {
      x -= MOD;
    }
    return *this;
  }
  mint operator++(int) {
    mint ret = x;
    if ((++x) >= MOD) {
      x -= MOD;
    }
    return ret;
  }
  mint& operator--() {
    if ((--x) < 0) {
      x += MOD;
    }
    return *this;
  }
  mint operator--(int) {
    mint ret = x;
    if ((--x) < 0) {
      x += MOD;
    }
    return ret;
  }
  mint operator+(const mint& r) {
    mint ret;
    return ret = x + r.x;
  }
  mint operator-(const mint& r) {
    mint ret;
    return ret = x - r.x;
  }
  mint operator*(const mint& r) {
    mint ret;
    return ret = x * r.x;
  }
  mint operator-() {
    return mint() - *this;
  }
  bool operator<(const mint& a) {
    return x < a.x;
  }
  bool operator>(const mint& a) {
    return x > a.x;
  }
  bool operator<=(const mint& a) {
    return x <= a.x;
  }
  bool operator>=(const mint& a) {
    return x >= a.x;
  }
  bool operator==(const mint& a) {
    return x == a.x;
  }
  bool operator!=(const mint& a) {
    return x != a.x;
  }
};
ostream& operator<<(ostream& os, const mint& r) {
  os << r.x;
  return os;
}
ostream& operator>>(ostream& os, const mint& r) {
  os >> r.x;
  return os;
}

void solve(istream& cin, ostream& cout) {
  int n, k;
  cin >> n >> k;
  // 10^100 の数, 余りのバリエーションを数える
  // a * 10^100 + b とし，考えられる (a,b) の組を考える
  mint ans = 0;
  for (ll a = k; a <= n + 1; a++) {
    ll min_b = a * (a - 1) / 2;
    ll max_b = (n * 2 - a + 1) * a / 2;
    ans += max_b - min_b + 1;
    // mint(10 * a + b)
  }
  cout << ans << endl;
}

#ifndef TEST
int main() {
  OPTIMIZE_STDIO;
  solve(cin, cout);
  return 0;
}
#endif