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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
typedef long long ll;
typedef long double ld;
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF (1LL<<60)
#define MOD1 1000000007

void swap(ll *a, ll *b){ll c;c=*b;*b=*a;*a=c;}/*swap 交換*/
ll max2(ll a,ll b){return a>=b?a:b;}/*2つのうち大きい数を返す*/ 
ll min2(ll a,ll b){return a>=b?b:a;}/*2つのうち小さい数を返す*/ 
ll ABS(ll a){return a>=0?a:(-a);}/*絶対値*/ 
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}

typedef struct{
    ll aa;
    ll bb;
}frequent;

int main(void){
    ll n,a,b,ans=0;
    scanf("%lld%lld%lld",&n,&a,&b);
    if(a==0){
        printf("0\n");
        return 0;
    }
    // rep(i,0,n){
    //     scanf("%lld",&[i]);
    // }
    ans+=n/(a+b)*a;
    n-=(n/(a+b)*a);
    ans+=n>a?n-b:n;
    printf("%lld\n",ans);
    return 0;
}