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 <bits/stdc++.h>
using namespace std;

//#define int long long
#define reps(i,s,n) for(int (i)=(s);(i)<(n);++(i))
#define rep(i,n) reps(i,0,n)
#define rept(i,n) reps(i,0,((n)+1))
#define repst(i,s,n) reps(i,s,((n)+1))
#define reprt(i,n,t) for(int (i)=(n);(i)>=(t);--(i))
#define repr(i,n) reprt(i,n,0)
#define each(itr,v) for(auto &(itr):(v))
#define all(c) (c).begin(),(c).end()
#define pb push_back
#define mp(x,y) make_pair((x),(y))
#define fi first
#define se second
#define tmax(x,y,z) max(x,max(y,z))
#define tmin(x,y,z) min(x,min(y,z))
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
#define ln '\n'
#define bln(i,n) (i==n-1?'\n':' ')
#define dbg(x) cout<<#x" = "<<(x)<<ln
#define dbga(x,n) {cout<<#x" : ";for(int (i)=0;i<(n);++i){cout<<((x)[i])<<(i==((n)-1)?'\n':' ');}}

typedef long double ld;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<string> vst;
typedef vector<bool> vb;
typedef vector<ld> vld;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<vector<int> > mat;

const ll inf = (ll)1e9+10;
const ll linf = (ll)1e18+10;
const ll mod = (ll)(1e9+7);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int ddx[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[] = {1, 1, 0, -1, -1, -1, 0, 1};
const double eps = 1e-10;

struct oreno_initializer {
	oreno_initializer() {
		cin.tie(0);
		ios::sync_with_stdio(0);
	}
} oreno_initializer;

// ━━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…
// .｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋.｡.:( ^ω^)・ﾟ＋
// ・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・‥…━━━☆・

// b==2) 1,3,5,7,...でk>=1
// b==3) 1,2, 4,5, 7,8,

int n, k;
ll res;

signed main() {
	cin >> n >> k;
	if (k==0) return cout << n*n << ln, 0;
	repst(b,k+1,n) {
		int cy = n/b;
		res += cy * (b-k);
		int st = b * cy, to = n;
		res += max(0, to%b - k + 1);
	}
	cout << res << ln;
}