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)

//    Unmesh Kumar
//    CSE'22 IIT Roorkee

#include <iostream>
#include <bits/stdc++.h>
#include <sstream>
#include <cmath>
#include <algorithm>
#include <map>
#include <iterator>
#include <iomanip>
#include <queue>
#include <utility>
#include <set>

using namespace std;
#define ll long long int
#define c(x) ll x;cin>>x
#define cc(x,y) ll x,y;cin>>x>>y
#define ccc(x,y,z) ll x,y,z; cin>>x>>y>>z
#define db long double
#define fast cin.tie(NULL); cout.tie(NULL); ios_base::sync_with_stdio(false)
#define inf 200000000000000
#define mod 1000000007
#define eps 1e-7
#define PI 3.1415926535897932385
#define pb push_back
#define bitc  __builtin_popcountll
#define mp make_pair
#define ff first
#define ss second
#define all(ar) ar.begin(), ar.end()
#define len(x) (ll)(x).size()
#define fr(i,a,b) for (ll i = (a), _b = (b); i <= _b; i++)
#define rep(i,n) for (ll i = 0, _n = (n); i < _n; i++)
#define repr(i,n) for (ll i = n - 1; i >= 0; i--)
#define frr(i,a,b) for (ll i = (a), _b = (b); i >= _b; i--)
#define foreach(it,ar) for (auto it = ar.begin(); it != ar.end(); it++)
#define fill(ar,val) memset(ar, val, sizeof(ar))
#define print(arr,n) rep(i,n) cout<<arr[i]<<" "; cout<<endl
#define print2(arr,n,m) rep(i,n){ rep(j,m) cout<<arr[i][j]<<" "; cout<<endl; }
#define dyn1(arr,n) ll* arr=new ll[n]; rep(i,n) arr[i]=0;
#define dyn2(arr,n,m) ll** arr=new ll*[(n)]; rep(i,n) arr[i]=new ll[m]; rep(i,n) rep(j,m) arr[i][j]=0
#define carr(arr,n) rep(i,n) cin>>arr[i]
#define carr2(arr,n,m) rep(i,n) rep(j,m) cin>>arr[i][j]
#define debug(a) cout<<" :: "<<#a<<" == "<<a<<endl
#define debug2(a,b) cout<<" :: "<<#a<<" == "<<a<<" :: "<<#b<<" == "<<b<<endl
typedef pair<int,int> ii;
typedef vector<ii> vii;

inline ll maxim(ll a,ll b) {if(a>b) return a; else return b;}
inline ll minim(ll a,ll b) {if(a<b) return a; else return b;}
inline bool equals(double a, double b){ return fabs(a - b) < 1e-9; }
ll gcd(ll a, ll b){ return b==0 ? a : gcd(b, a%b); }


ll binpow(ll a, ll b)
{
    ll res = 1;
    while (b > 0)
    {
        if (b & 1) res = (res * a)%mod;
        a = (a * a)%mod;
        b >>= 1;
    }
    return res;
}

int modInverse(int a, int m)
{
    int m0 = m;
    int y = 0, x = 1;
    if (m == 1)
      return 0;
    while (a > 1)
    {
        int q = a / m;
        int t = m;
        m = a % m, a = t;
        t = y;
        y = x - q * y;
        x = t;
    }
    if (x < 0)
       x += m0;
    return x;
}

bool comp(ll a,ll b)
{
  return a>b;
}

ll arr[1000005];
ll a[1000005];
ll b[1000005];

int main()
{
  fast;
  //freopen("input.txt", "r", stdin);
  //freopen("output.txt", "w", stdout);
  cout<<setprecision(10)<<fixed;
  c(n);
  cc(a,b);
  ll total=0;
  total+=(n/(a+b))*a;
  ll res=n%(a+b);
  if(res>=a) total+=a;
  else total+=res;
  cout<<total<<endl;

  return 0;
}