Problem:
Score : 200 points

Problem Statement

Takahashi has many red balls and blue balls. Now, he will place them in a row.

Initially, there is no ball placed.

Takahashi, who is very patient, will do the following operation 10^{100} times:

Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.

How many blue balls will be there among the first N balls in the row of balls made this way?

Constraints

1 \leq N \leq 10^{18}

A, B \geq 0

0 < A + B \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the number of blue balls that will be there among the first N balls in the row of balls.

Sample Input 1

8 3 4

Sample Output 1

4

Let b denote a blue ball, and r denote a red ball. The first eight balls in the row will be bbbrrrrb, among which there are four blue balls.

Sample Input 2

8 0 4

Sample Output 2

0

He placed only red balls from the beginning.

Sample Input 3

6 2 4

Sample Output 3

2

Among bbrrrr, there are two blue balls.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
using ll = int64_t;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using mii = map<int, int>;
using vi = vector<int>;
using vii = vector<vi>;
using vl = vector<ll>;
using vll = vector<vl>;
using tpi = tuple<int, int, int>;

using vqi = vector<queue<int>>;
using vpi = vector<pii>;
using bs = bitset<60>;
using vbi = vector<bs>;

ll gcd(ll a, ll b) {
    if (a < b) {
        swap(a, b);
    }

    /* ユークリッドの互除法 */
    ll r = a % b;
    while (r != 0) {
        a = b;
        b = r;
        r = a % b;
    }
    return b;
}

ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

constexpr int64_t __pow_mod(int64_t n, int64_t p, int64_t mod) noexcept {
    if (p == 0) {
        return 1;
    }
    if (p % 2 == 0) {
        int64_t t = __pow_mod(n, p / 2, mod);
        return (t * t) % mod;
    }
    return (n * (__pow_mod(n, p - 1, mod) % mod)) % mod;
}

constexpr int64_t pow_mod(int64_t n, int64_t p, int64_t mod) noexcept {
    int64_t res = __pow_mod(n, p, mod);
    if (res == 0) {
        res = mod;
    }
    return res;
}

constexpr int64_t nCr_mod(int64_t n, int64_t r, int64_t mod) noexcept {
    int64_t count = 1;
    auto a = r;
    for (auto i = n; i > n - r; i--) {
        count *= i;
        if (count % mod != 0) {
            count %= mod;
        }
        int64_t inv = pow_mod(a, mod - 2, mod);
        count *= inv;
        if (count % mod != 0) {
            count %= mod;
        }
        a--;
    }
    return count;
}

void solve(long long N, long long A, long long B) {
    if (A >= N) {
        cout << N << endl;
        return;
    }

    if (A + B >= N) {
        cout << A << endl;
        return;
    } else if (A + B == 0) {
        cout << 0 << endl;
        return;
    } else {
        ll ans = A * (N / (A + B));
        ll m = N % (A + B);
        if (m >= A) {
            ans += A;
        } else {
            ans += m;
        }
        cout << ans << endl;
        return;
    }
}

int main() {
    long long N;
    std::cin >> N;
    long long A;
    std::cin >> A;
    long long B;
    std::cin >> B;
    solve(N, A, B);
    return 0;
}