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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>


int compare_int(const void *a, const void *b)
{
//    return *(int*)a - *(int*)b;
    return *(long long*)a - *(long long*)b;
}
//  qsort(A, b, sizeof(int), compare_int);

//long long  factorial(long long n) {
//    if (n > 0) {
//        return n * factorial(n - 1);
//    } else {
//        return 1;
//    }
//}
long long  factorial(long long n, long long depth) {
    if (n > 1) {
        depth++;
        return factorial(n / 2, depth);
    } else {
        return depth;
    }
}

int min(int a, int b) {
    if (a > b)
        return b;
    else
        return a;
}

int distance(int a, int b) {
  return abs(a-b)*abs(a-b);
}



//int ()
/*
int triangle(int a, int b, int c){
  if(a<b+c){
    if(b<c+a){
      if(c<a+b){
        printf("1\n");
        return 1;
      }
    }
  }
  printf("1\n");
  return 0;
}
*/

int main()
{
//  int  a ,b, c, d, e, X, Y ,r;
  unsigned long long  a, b, c, d, e, X, Y;
  //  double  c;
// long long  A[200003]={};
  int  A[100003]={};
  int  B[100003]={};
//  int  C[100003][100001]={};
  int  C[100003]={};

  int  flg=0;

//  char s[101][101]={};
  char s[100001]={};
  char t[100001]={};
//  int N, M;
//  int A[1000000001]={};
//  long long  C[101]={};
  int i,j,k;
  long long p=0,q=0;
//  int flag=0;
//  int tmp=0;
//  char tmp[101]={};
//  long long min=10000000000;
//  long long min=100000001;
  long long min =100000001;
  int len=0;
  int l=0;
//  int r=1000001;
  int cnt=0;
  int num=0;
  int num1=0;
  int num2=0;
  long long  ans=0;
  int ans_ac=0;
  int ans_wa=0;
  int max=0;
  long long sum=0;
  int addr=0;
  int ptn=0;
  int  tmp=0;
  double per=0; 
//  bool on_off[100]={};
  
//  char alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

  //  a=0;
  scanf("%lld%lld%lld", &a,&b,&c);
  d=a/(b+c);
  
  X=a-((b+c)*(d));
//  printf("%lld\n", (b*(j))+X);
  if(X<=b)  printf("%lld\n", (b*(d))+X);
  else printf("%lld\n", (b*(d+1)));
  
  
return 0;
}