#include <bits/stdc++.h>
#define endl '\n'
using namespace std;

int n, k, cnt = 0;
long long dp[70010];

int main(){
	cin >> n >> k;
	for(int i = 1; i <= n; i++){
		dp[i] = dp[i-1] + i;
	}
	
	for(int i = k; i <= n; i++){
		long long curr = dp[i] - dp[i-k];
		// 判断这个数是不是某一个数的平方 
		if((int)sqrt(curr) * (int)sqrt(curr) == curr){
			cnt++;
		}
	}
	cout << cnt;
	return 0;
}