#include <bits/stdc++.h>
using namespace std;
int a[100005];
int main(){
	int n, k, c = 0;
	cin >> n >> k;
	for (int i = 1; i <= n; i++) {
		a[i] = a[i - 1] + i;
	}
	for (int i = k; i <= n; i++) {
		int temp = a[i] - a[i - k];
		if (sqrt(temp) == (int)sqrt(temp)) {
			c++;
		}
	}
	cout << c;
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
int a[100005];
int main(){
	int n, k, c = 0;
	cin >> n >> k;
	for (int i = 1; i <= n; i++) {
		a[i] = a[i - 1] + i;
	}
	for (int i = k; i <= n; i++) {
		int temp = a[i] - a[i - k];
		if (sqrt(temp) == (int)sqrt(temp)) {
			c++;
		}
	}
	cout << c;
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
long long a[70005];
int main(){
    long long s=0;
    long long n,k;
	cin>>n>>k;
	for(int i=1;i<=n;i++){
		a[i]=a[i-1]+i;//构建前缀和
	}
	for(int i=k;i<=n;i++){//循环k范围的最后一个位置
		int x=a[i]-a[i-k];//找到k区间内的和
        if(sqrt(x)==(int)sqrt(x)){//区间和是完全平方数
	    	s++;
	    } 
	}
	cout<<s;
}