F

Mr. Potato Head has been promoted and now is a math professor at the UNAL.

For his first course he is willing to teach hard subjects, so at the moment he is teaching how to add and subtract fractions.

To complete his course the students have to do a long series of exercises, each exercise corresponds to a valid formula containing only additions and subtractions of fractions.

Formally a valid formula is one of the following:

• A fraction
• F1+F2F1+F2
• F1−F2F1−F2
• (F1)(F1)

where F1F1 and F2F2 are also valid formulas.

Mr. Potato Head knows that the exam would be impossible if fractions are too large or if they are negative, so he decides that for every fraction a/ba/b, 0≤a≤1000≤a≤100 and 0<b≤200<b≤20.

Can you pass the course of Mr. Potato Head?

Input

The input consists of several lines, each line contains a valid formula without spaces.

It is guaranteed that all lines contains valid formulas and the total number of characters in all formulas does not exceed 2∗1052∗105

Output

For each formula output a line with an irreducible fraction a/ba/b, b>0b>0 −− The solution of the corresponding formula

Example

Input

1/2+1/3
1/5-2/10
1/2+(1/2-2/1)

Output

5/6
0/1
-1/1

Note

A fraction is irreducible if its numerator and denominator do not have common divisors greater than 1

题解:只有加减法和括号的运算,可直接从前往后挨个计算,只需判断当前的符号是+或-即可.

#include <bits/stdc++.h>
#define met(a, b) memset(a, b, sizeof(a))
#define ll long long
#define ull unsigned long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
typedef pair<ll,ll>P;
const int maxn=200010;
const double eps=1e-10;
const double pi=acos(-1);
char s[maxn];
int len;
P ans;
int f,ff;
P cal(P ans,P now)
{
  ll son=ans.first*now.second+f*ff*now.first*ans.second;
  ll mon=ans.second*now.second;
  ll g=gcd(abs(son),abs(mon));
  son/=g;
  mon/=g;
  return P(son,mon);
}
P ok(int pos)
{
  ll son=0,mon=0,l=-1;
  for(int i=pos-1;i>=0;i--){
    if(s[i]<'0'||s[i]>'9'){
      l=i;
      break;
    }
  } 
  for(int i=l+1;i<len;i++){
    if(s[i]>='0'&&s[i]<='9')son=son*10+s[i]-'0';
    else break;
  }
  for(int i=pos+1;i<len;i++){
    if(s[i]>='0'&&s[i]<='9')mon=mon*10+s[i]-'0';
    else break;
  }
  return P(son,mon);
}
int main()
{
  while(cin>>s){
    stack<int>sta;
    len=strlen(s);
    ans=P(0,1);
    f=ff=1;
    for(int i=0;i<len;i++){
      if(s[i]=='+'){
        f=1;
      }
      else if(s[i]=='-'){
        f=-1;
      }
      else if(s[i]=='('){
        f=1;
        if(s[i-1]=='-'){
          ff*=-1;
          sta.push(-1);
        }
        else{
          sta.push(1);
        }
      }
      else if(s[i]==')'){
        int w=sta.top();
        ff*=w;
        sta.pop();
      }
      else if(s[i]=='/'){
        P now=ok(i);
        ans=cal(ans,now);
      }
    }
    cout<<ans.first<<"/"<<ans.second<<endl;
  }
  return 0;
}