Problem:
Score : 400 points

Problem Statement

Takahashi had a pair of two positive integers not exceeding N, (a,b), which he has forgotten.
He remembers that the remainder of a divided by b was greater than or equal to K.
Find the number of possible pairs that he may have had.

Constraints

1 \leq N \leq 10^5

0 \leq K \leq N-1

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of possible pairs that he may have had.

Sample Input 1

5 2

Sample Output 1

7

There are seven possible pairs: (2,3),(5,3),(2,4),(3,4),(2,5),(3,5) and (4,5).

Sample Input 2

10 0

Sample Output 2

100

Sample Input 3

31415 9265

Sample Output 3

287927211

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*  -*- coding: utf-8 -*-
 *
 * b.cc: D: Remainder Reminder
 */

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;

/* constant */

/* typedef */

typedef long long ll;

/* global variables */

/* subroutines */

/* main */

int main() {
  int n, k;
  scanf("%d%d", &n, &k);

  if (k == 0) {
    printf("%lld\n", (ll)n * n);
    return 0;
  }
  
  ll sum = 0;
  for (int b = k + 1; b <= n; b++) {
    int c = n / b, d = n % b;
    sum += (ll)c * (b - k) + (d >= k ? (d + 1 - k) : 0);
  }

  printf("%lld\n", sum);
  return 0;
}