Problem:
Score : 300 points

Problem Statement

Compute A \times B, truncate its fractional part, and print the result as an integer.

Constraints

0 \leq A \leq 10^{15}

0 \leq B < 10

A is an integer.

B is a number with two digits after the decimal point.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the answer as an integer.

Sample Input 1

198 1.10

Sample Output 1

217

We have 198 \times 1.10 = 217.8. After truncating the fractional part, we have the answer: 217.

Sample Input 2

1 0.01

Sample Output 2

0

Sample Input 3

1000000000000000 9.99

Sample Output 3

9990000000000000

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

/**
 *    +--+--+--++--++--In the name of ALLAH--++--++--+--+--+
 *
 *    author:  skmonir
 *    created: 31-May-2020 18:07:13
**/

#include <bits/stdc++.h>

using namespace std;

#define endl                '\n'
#define TN                  typename
#define mod                 1000000007LL
#define len(x)              (int) x.size()
#define all(x)              x.begin(), x.end()
#define rall(x)             x.rbegin(), x.rend()
#define mem(x, y)           memset(x, y, sizeof x)
#define FOR(x, l, r)        for (int x = l; x <= r; ++x)
#define ROF(x, l, r)        for (int x = l; x >= r; --x)

template <TN T> inline void Int(T &n) {
    n = 0; int f = 1; int ch = getchar();
    for (; !isdigit(ch); ch = getchar()) if (ch == '-') f = -1;
    for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
    n = n * f;
}

template <TN T> T gcd(T a, T b) {return !b ? a : gcd(b, a % b);}
template <TN T> inline void umin(T &a, T b) {a = a < b ? a : b;}
template <TN T> inline void umax(T &a, T b) {a = a > b ? a : b;}
template <TN T, TN W> inline void Int(T &x, W &y) {Int(x), Int(y);}
template <TN T, TN W, TN Q> inline void Int(T &x, W &y, Q &z) {Int(x, y), Int(z);}

const int N = 1e5 + 7;
const int inf = 1e9 + 7;


int solve() {
    long long A;
    double b;
    cin >> A >> b;

    b *= 1000.0;

    long long B = (long long) b;
    B /= 10;

    long long c = A * B;

    cout << c / 100 << endl;

    return 0;
}

int main() {
    int tests = 1, CaseNo = 0;
    // Int(tests);
    while (tests--) {
        //printf("Case %d: ", ++CaseNo);
        solve();
    }
    return 0;
}