#include <iostream>
#include <cmath>
using namespace std;
typedef long long int ll ;
ll mod = 1000000007;
int main() {
ll b ;
ll a = 1;
ll c ;
ll d;
ll n, k ;
cin >> n >> k ;
b = -((2*(n%mod))%mod + 3);
//cout << b ;
c = -(k - n*n - n );
b = b%mod;
d = (b*b)%mod - (4*1*(c)%mod)%mod ;
long double ans = (-b - sqrtl(d))/(2*1) ;
if (ans < 0 )
{
ans = (-b + sqrtl(d))/(2*1) ;
}
cout << (ans) << endl ;
return 0;
}
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