Problem: Using the given four integers from 1 to 9, we create an expression that gives an answer of 10. When you enter four integers a, b, c, d, write a program that outputs an expression that gives an answer of 10 according to the following conditions. Also, if there are multiple answers, only the first answer found will be output. If there is no answer, output 0.

* Use only addition (+), subtraction (-), and multiplication (*) as operators. Do not use division (/). You can use three operators.
* You must use all four numbers.
* You can freely change the order of the four numbers.
* You can use parentheses. You can use up to 3 sets (6) of parentheses.



Input

Given multiple datasets. The format of each dataset is as follows:


a b c d

Input ends with four 0s. The number of datasets does not exceed 40.

Output

For each dataset, combine the given four integers with the above arithmetic symbols and parentheses to output an expression or 0 with a value of 10 on one line. The expression string must not exceed 1024 characters.

Example

Input

8 7 9 9
4 4 4 4
5 5 7 5
0 0 0 0


Output

((9 * (9 - 7)) - 8)
0
((7 * 5) - (5 * 5))

Solution (in CPP):

Solution:
#include <cstdio>
#include <utility>
#include <cctype>
#include <array>
#include <algorithm>
#include <functional>
using namespace std;

#ifdef __linux
#define _U(s) s##_unlocked
#else
#define _U(s) _##s##_nolock
#define _CRT_DISABLE_PERFCRIT_LOCKS
#endif
#define gcu _U(getchar)
#define pcu _U(putchar)
#define _DEF(r, n, ...) inline r n(__VA_ARGS__) noexcept
#define _T template <typename T>
#define _HT template <typename H,typename... T>
#define _OP(t) inline operator t()
struct _in {
#ifdef _GLIBCXX_STRING
	_OP(string){string s;char c;while(isspace(c = gcu()));do{s+=c;}while(c=gcu(),c!=' '&&c!='\n'&&c!=EOF);return s;}
#define _S
#endif
	_OP(char){char c=gcu();gcu();return c;}
	_OP(double){double d; scanf("%lf",&d); gcu();return d;}
	_T _OP(T){T n{},m{1},c;while(isspace(c = gcu()));if(c=='-')m=-1,c=gcu();do{n=10*n+(c-'0'),c=gcu();}while(c>='0'&&c<='9');return m*n;}
} in;
#define _SCAN(...) _DEF(bool,scan,__VA_ARGS__)
#ifdef _S
_SCAN(string &o) {int c{gcu()};if(c==EOF)return false;else{ungetc(c,stdin);string t=move(in);o=t;return true;}}
#endif
_T _SCAN(T &o) {int c{gcu()};return c==EOF?false:(ungetc(c,stdin),o=in,true);}
_HT _SCAN(H &h,T&&... t){return scan(h)&&scan(t...);}
#define _OUT(...) _DEF(void,out,__VA_ARGS__)
#define _OUTL(...) _DEF(void,outl,__VA_ARGS__)
_OUT(bool b){pcu('0'+b);}
_OUT(const char *s){while(*s)pcu(*s++);}
_OUT(char c){pcu(c);}
#ifdef _S
_OUT(string &s){for(char c:s)pcu(c);}
#endif
_T _OUT(T n){static char b[20];char *p=b;T m=n<0?pcu('-'),-1:1;
	if(!n)*p++='0';else while(n)*p++=(char)(n%10*m+'0'),n/=10;while(p!=b)pcu(*--p);}
_OUTL(){out('\n');}
#ifdef _GLIBCXX_VECTOR
	_T _OUT(vector<T> &v){for(T &x:v)out(&x == &v[0]?"":" "),out(x);}
#endif
_HT _OUT(H &&h, T... t){out(h);out(t...);}
template <typename... T> _OUTL(T... t){out(t...);outl();}
struct range{
	int e,b=0,s=1;range(int _b,int _e,int _s):e(_e),b(_b),s(_s){} range(int _b,int _e): e(_e), b(_b){} range(int _e):e(_e){}
	struct it{int v,s; it(int _v,int _s):v(_v),s(_s){} operator int()const{return v;} operator int&(){return v;}int operator*()const{return v;}
		it& operator++(){v+=s;return *this;} }; it begin(){return {b,s};} it end(){return {e,s};}};
#define times(i,n) for(int i=n;i;i--)

char op(int i) {
	return "+-*"[i];
}

int main() {
	function<int(int, int)> f[] = {plus<int>(), minus<int>(), multiplies<int>()};
	range r3(3);
	for (array<int, 4> a; a = {in, in, in, in}, a[0];)
		[&]() {
			sort(begin(a), end(a));
			do {
				for (int i: r3)
					for (int j: r3)
						for (int k: r3) {
							auto fmt = [&](const char *f) {
								printf(f, a[0], op(i), a[1], op(j), a[2], op(k), a[3]);
								outl();
							};
							if (f[i](a[0], f[k](f[j](a[1], a[2]), a[3])) == 10)
								return fmt("(%d %c ((%d %c %d) %c %d))");
							if (f[i](a[0], f[j](a[1], f[k](a[2], a[3]))) == 10)
								return fmt("(%d %c (%d %c (%d %c %d)))");
							if (f[k](f[j](f[i](a[0], a[1]), a[2]), a[3]) == 10)
								return fmt("(((%d %c %d) %c %d) %c %d)");
							if (f[k](f[i](a[0], f[j](a[1], a[2])), a[3]) == 10)
								return fmt("((%d %c (%d %c %d)) %c %d)");
							if (f[j](f[i](a[0], a[1]), f[k](a[2], a[3])) == 10)
								return fmt("((%d %c %d) %c (%d %c %d))");
						}
			} while (next_permutation(begin(a), end(a)));
			outl(0);
		}();
}