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;
#define pi acos(-1.0)
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define rep(i,a,n) for (ll i=a;i<n;i++)
#define per(i,a,n) for (ll i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((ll)(x).size())
#define cin1(x) cin >> x
#define cin2(x, y) cin >> x >> y
#define cin3(x, y, z) cin >> x >> y >> z
#define cin4(x, y, z, _) cin >> x >> y >> z >> _
#define cin5(x, y, z, p, q) cin >> x >> y >> z >> p >> q
#define cou(x) cout << x
#define cous(x) cout << x << " "
#define coul(x) cout << x << "\n"
#define ret(x) return x
#define MAX 5000001
#define start int main() {   fastio
#define finish ret(0); }
#define Test ll _t cin1(_t); rep(i, 0, _t)

struct point{    int x, y, ind;   char ch;  };
typedef long long ll;
typedef vector<ll> VI;
typedef pair<ll,ll> PII;
typedef double db;
//mt19937 mrand(random_device{}());
const ll mod=1000000007;
//ll rnd(ll x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}ret(res);}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
ll fact[13];
void factorial() {   fact[0]=1;  rep(i, 1, 12) fact[i]=(fact[i-1]*i);  }
vector <ll> sie; //for sieve function
ll spf[MAX+7];  //for sieve function
void sieve() {    rep(i, 2, MAX)  {  if (spf[i] == 0) {  spf[i] = i;   sie.pb(i);  }    for (ll j=0; j<SZ(sie) && i*sie[j] <= MAX && sie[j] <= spf[i]; j++)   spf[i * sie[j]] = sie[j]; }   }
bool cmp(point A, point B){   if(A.x != B.x) ret(A.x < B.x);  ret(A.y < B.y);   }
// head collected

ll n, a, b, re, ans, ex;
string str;
char ara_a[55][55];
char ara_b[55][55];

start;

    cin3(n, a, b);
    re = n / (a + b);
    ans = re * a;
    ex = n % (a + b);
    if ( ex > a )
        ex = a;
    ans += ex;

    coul(ans);


finish