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 = long long;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repe(i, n) for (int i = 0; i <= (int)(n); i++)
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define ANS cout << ans << endl
#define all(a) a.begin(),a.end()
#define SORT(a) sort(all(a))
#define REVERSE(a) reverse(all(a))
#define KOZYUN(a) sort(all(a),greater<int>())
#define int ll
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORe(i, m, n) for(int i = m;i <= n;i++)
#define VIin(a) for(int i=0; i<a.size(); i++) cin >> a[i]
#define OUT(a) cout << a << endl
#define IN(a) cin >> a
#define INT(a) int a; cin >> a
typedef vector<int> VI;
typedef vector<VI> VVI;
const int MOD=1e9+7;
const long long INF = 1LL << 60;
/*最大公約数*/int gcd(int x, int y) { return y ? gcd(y,x%y) : x;}
/*最小公倍数*/int lcm(int a, int b) { int g = gcd(a, b); return a / g * b; }
/*素数判定*/int isPrime(int n){if(n<2){return 0;}for(int i=2; i*i<=n; i++){if(n%i==0){return 0;}}return 1;}
/*約数列挙*/vector<int> getDivisor(int n){vector<int> v; for(int i=1; i*i<=n; i++){if(n%i==0){v.push_back(i);if(i!=n/i){v.push_back(n/i);}}}sort(v.begin(),v.end()); return v;}
//----------------------------------------------
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n,b,a; cin >> n >> a >> b;
    if(a == 0){OUT(0); return 0;}
    int ans = 0;
    ans += n/(a+b) * a;
    int x = n%(a+b);
    if(x < a) ans += x;
    else ans += a;
    ANS;

    return 0;
}