#include <bits/stdc++.h>

#define SIZE 300005
#define MOD 1000000007LL
#define INF INT_MAX / 3
#define LLINF 1LL << 60
#define REP(i,n) for(int i=0;i<n;i++)
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define DOWN(i,b,a) for(int i=b;i>=a;i--)
#define SET(a,c) memset(a,c,sizeof a)
#define FORALL(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define FOREACH(i,c) for(auto (i) : (c))
#define BIT(i,j) ((i)>>(j))&1
#define ALL(o) (o).begin(), (o).end()
#define ERASE(o) (o).erase(unique((o).begin(),(o).end()), (o).end())
#define SQ(x) ((x)*(x))
using namespace std;
typedef long long ll;
typedef valarray<int> Array;
typedef pair<ll,ll> Pll;
typedef pair<int, int> Pii;
typedef pair<double, double> Pdd;
template<typename T> inline void priv(vector<T>a){REP(i,a.size()){cout<<a[i];if(i==a.size()-1)cout<<endl;else cout<<" ";}}
int gcd(int a,int b){int c=max(a,b);int d=min(a,b);return c==0||d==0?c:gcd(c%d,d);}
int lcm(int a,int b){return a==0||b==0?0:a*b/gcd(a,b);}

map<string, int> generate(vector<int> v)
{
  map<string, int> T;
  if(v.size() == 1)
  {
    if(v[0] < 0) T["("+to_string(v[0])+")"] = v[0];
    else T[to_string(v[0])] = v[0];
    return T;
  }
  else
  {
    REP(i,v.size()-1)
    {
      vector<int> a,b;
      map<string, int> s,t;
      FOR(j,0,i) a.push_back(v[j]);
      FOR(j,i+1,v.size()-1) b.push_back(v[j]);
      s = generate(a);
      t = generate(b);
      for(auto m : s) for(auto n : t)
      {
        T["("+m.first+" * "+n.first+")"] = m.second * n.second;
        T["("+m.first+" + "+n.first+")"] = m.second + n.second;
        T["("+m.first+" - "+n.first+")"] = m.second - n.second;
      }
    }
    return T;
  }
}

void solve(vector<int> v)
{
  do
  {
    map<string, int> T = generate(v);
    for(auto a : T) if(a.second == 10)
    {
      printf("%s\n",a.first.c_str());
      return;
    }
  } while(next_permutation(v.begin(),v.end()));
  printf("0\n");
}

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  vector<int> a(4);
  // FOR(i,1,9)FOR(j,1,9)FOR(k,1,9)FOR(l,1,9)
  // {
  //   a[0] = i; a[1] = j; a[2] = k; a[3] = l;
  //   solve(a);
  // }
  while(1)
  {
    REP(i,4) cin >> a[i];
    if(!accumulate(a.begin(),a.end(),0)) break;
    sort(a.begin(),a.end()); solve(a);
  }
  return 0;
}