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: Wrong Answer)

#include <iostream>
#include <fstream>
#include <string> 
#include <cmath>  
#include <cstdlib>
#include <ctime>
#include <vector>
#include <algorithm>
#include <numeric>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <functional>

using namespace std;
using ll = long long;

#define FOR(i, m, n) for(int i = (m);i < (n);i++)
#define REP(i,n) for(int i = 0; i < int(n); i++)
#define VI vector<int>
#define VVI vector<vector<int>>
#define VVVI vector<vector<vector<int>>>
#define VL vector<ll>
#define VVL vector<vector<ll>>
#define VB vector<bool>
#define VVB vector<vector<bool>>
#define PAIR pair<int,int>
#define MP make_pair
#define VP vector<pair<int,int>>
#define VS vector<string>
#define QUE queue<int>
#define DEQ deque<int>
#define PQUE priority_queue<int> //5,5,4,3,3,2,...
#define REPQUE priority_queue<int, vector<int>, greater<int>> //1,1,2,3,4,4,5,...
#define SUM(obj) accumulate((obj).begin(), (obj).end(), 0)
#define SORT(obj) sort((obj).begin(), (obj).end()) // 1,2,3,4,5...
#define RESORT(obj) sort((obj).begin(), (obj).end(), greater<int>()) // 5,4,3,2,1...
#define UB(obj,n) upper_bound((obj).begin(), (obj).end(), n) //itr > n
#define LB(obj,n) lower_bound((obj).begin(), (obj).end(), n) //itr>= n

const ll MOD = (ll)1e9 + 7;

int main() {
	int N, K;
	cin >> N >> K;
	ll sum = 0;
	FOR(i, 1, N + 1) {
			sum += (N / i)*max(0, i - K);
			sum += max(0,N % i - K + 1);
	}
	cout << sum << endl;
	//cin >> N >> K;


	
	return 0;
}