#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using ll = long long;
void solve() {
  string s;
  cin >> s;
  int m = ((int)(s).size());
  vi a(m);
  a[0] = ((s[0] == '-') ? -1 : 1);
  for (__typeof(m) i = (1) - ((1) > (m)); i != (m) - ((1) > (m));
       i += 1 - 2 * ((1) > (m))) {
    if (s[i] == '-') {
      a[i] = a[i - 1] - 1;
    } else {
      a[i] = a[i - 1] + 1;
    }
  }
  ll res = m;
  int i = 0, t = -1;
  while (i < m) {
    if (a[i] != t) {
      i++;
    } else {
      t--;
      res += (i + 1);
    }
  }
  cout << res << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode: 
    
    
    res = 0  
    for init = 0 to inf  
        cur = init  
        ok = true  
        for i = 1 to |s|  
            res = res + 1  
            if s[i] == '+'  
                cur = cur + 1  
            else  
                cur = cur - 1  
            if cur < 0  
                ok = false  
                break  
        if ok  
            break  
    

Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.

You have to calculate the value of the res after the process ends.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.

It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.

Output

For each test case print one integer — the value of the res after the process ends.

Example

Input


3
--+-
---
++--+-


Output


7
9
6

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using ll = long long;
void solve() {
  string s;
  cin >> s;
  int m = ((int)(s).size());
  vi a(m);
  a[0] = ((s[0] == '-') ? -1 : 1);
  for (__typeof(m) i = (1) - ((1) > (m)); i != (m) - ((1) > (m));
       i += 1 - 2 * ((1) > (m))) {
    if (s[i] == '-') {
      a[i] = a[i - 1] - 1;
    } else {
      a[i] = a[i - 1] + 1;
    }
  }
  ll res = m;
  int i = 0, t = -1;
  while (i < m) {
    if (a[i] != t) {
      i++;
    } else {
      t--;
      res += (i + 1);
    }
  }
  cout << res << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```