Maximize expression

//https://www.spoj.com/problems/LISA/

// I can't tell you what it really is,
// I can only tell you what it feels like.
#include "bits/stdc++.h"
using namespace std;
#define int long long

auto operator+=(pair<int,int> &x, const pair<int,int> &y) {
  x.first += y.first;
  x.second += y.second;
  return x;
}
auto operator*=(pair<int,int> &x, const pair<int,int> &y)  {
  x.first *= y.first;
  x.second *= y.second;
  return x;
}
auto operator-=(pair<int,int> &x, const pair<int,int> &y)  {
  x.first -= y.first;
  x.second -= y.second;
  return x;
}

const int N = 102;
const int inf = numeric_limits<int>::max();
const int ninf = numeric_limits<int>::min();
char s[N];
int a[N];
pair<int,int> dp[N][N];
const pair<int,int> ok = make_pair(-1,-1);

void pre() {
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < N; ++j) {
      dp[i][j] = ok;
    }
  }
}

pair<int,int> go(int l, int r) {
  if (l == r) return make_pair(a[l],a[l]);
  if (dp[l][r] != ok) return dp[l][r];
  int ans = inf;
  pair<int,int> findL, findR;
  int mn = inf, mx = ninf;
  for (int i = l; i < r; ++i) {
    findL = go(l,i);
    findR = go(i+1,r);
    switch(s[i]) {
      case '+':
        findL += findR; break;
      case '*':
        findL *= findR; break;
      case '-':
        findL -= findR; break;
    }
    mn = min(mn, findL.first);
    mx = max(mx ,findL.second);
  }
  return dp[l][r] = make_pair(mn, mx);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string str;
  int t = 1; cin >> t;
  while (t--) {
    int n = 0;
    pre();
    cin >> str;
    for (int i = 0; i < str.size(); ++i) {
      if (isdigit(str[i])) a[n++] = str[i] - '0';
      else s[n-1] = str[i];
    }
    auto it = go(0,n-1);
    cout << it.second << ' ' << it.first << '\n';
  }
  return 0;
}
// 1). Try forming/rearranging the equations if any
// 2). If problem asks to count, try counting in reverse(it can be easier sometimes)