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// Lupus Nocawy 22 XI 2015, PA2016
// http://potyczki.mimuw.edu.pl/
// Runda 2
// Zadanie: JED
// Jedynki [B]

#include <cstdio>
#include <iostream>
#include <cassert>
#include <cstring>
#include <algorithm>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define REP(i,n) for(int i=0, _n=n; i<_n; ++i)
#define FOR(i,a,b) for(int i=(a), _b=(b); i<=_b; ++i)
#define FORD(i,a,b) for(int i=(a), _b=(b); i>=_b; --i)
#define IT(i,x) __typeof__(x) i=x
#define FOREACH(it,x) for(__typeof__((x).begin()) it=(x).begin(); it!=(x).end(); ++it)
#define ALL(x) x.begin(),x.end()
#define MP make_pair
#define PB push_back
#define DEBUG(x) if(debug) cout << x << endl;
typedef long long int lli;
typedef unsigned long long int llu;
typedef pair<int,int> pii;
const int debug=0;

const int INF   = 2147483647;
const int max_k = 1000000000;	// 10^9

// The 3,500th prime is 32,609, it's square is 1,063,346,881 > 10^9
const int max_prime = 32609;

int sito[max_prime+1];
vector<int> prime;		// 2, 3, 5, 7, 11, 13, ...
void findPrimes(void){
	prime.reserve(max_prime/10);	// gestosc przy 10^6 to ok 7.8% i maleje
	prime.push_back(2);
	for(int i=3; i<=max_prime; i+=2){
		if(sito[i]==0){
			prime.push_back(i);
			for(int x=i*2; x<=max_prime; x+=i)
				sito[x]=i;
		}
	}
}

vector<int> factorize(int k) {
	vector<int> factors;
	int i=0;
	int p=prime[i];
	while(p*p <= k){
		while(k%p==0){
			factors.push_back(p);
			k/=p;
		}
		p=prime[++i];
	}
	if(factors.size()==0 || k>1)
		factors.push_back(k);
	return factors;
}

// int solutionLength[max_prime+1];
// string solutionString[max_prime+1];
// vector<int> factors[max_prime+1];
map<int,int> solutionLength;
map<int,string> solutionString;
map<int, vector<int> > factors;

pair<int,string> find(int k, bool doRecurse){
	// printf("find: %d\n", k);

	if(solutionLength.count(k))
		return make_pair(solutionLength[k], solutionString[k]);

	int length=0;
	string str="";

	if(factors[k].size()==0)
		factors[k] = factorize(k);

	REP(i,factors[k].size()) {
		int p = factors[k][i];
		if(solutionLength.count(p)) {
			length += solutionLength[p];
			if(str.length()==0)
				str = solutionString[p];
			else
				str += "*" + solutionString[p];
		}
		else {
			pair<int,string> S1 = find(p-1, 1);
			length += S1.first + 1;
			if(str.length()==0)
				str = "(" + S1.second + "+1)";
			else
				str += "*(" + S1.second + "+1)";
		}
	}

	solutionLength[k] = length;
	solutionString[k] = str;

	// printf("%d %s\n", length, str.c_str());

	if(doRecurse) {
		find(k-1, 0);
		if (length > solutionLength[k-1]+1) {
			solutionLength[k] = solutionLength[k-1]+1;
			solutionString[k] = "("+solutionString[k-1]+"+1)";
		}
	}

	return make_pair(solutionLength[k], solutionString[k]);
}

void solve(void){
	int k;
	scanf("%d ", &k);

	if(k==1){
		printf("1\n");
		return;
	}

	pair<int,string> solution = find(k,1);
	printf("%s\n", solution.second.c_str());

	// printf("%d %s\n", solution.first, solution.second.c_str());
	// printf("%d =%s\n", solution.first, solution.second.c_str());
	// printf("%d\n", solution.first);
	// printf("\n");

	return;
}

int main(void){
	int t;
	scanf("%d ", &t);

	findPrimes();

	solutionLength[2] = 2;
	solutionString[2] = "(1+1)";

	solutionLength[3] = 3;
	solutionString[3] = "(1+1+1)";

	solutionLength[4] = 4;
	solutionString[4] = "(1+1+1+1)";

	solutionLength[5] = 5;
	solutionString[5] = "(1+1+1+1+1)";

	while(t--)
		solve();
	return 0;
}