#include <bits/stdc++.h>
using namespace std;
string str;
const int tt = 1;
void solve() {
  long long int m, n, a, b, c;
  cin >> str;
  n = str.size();
  long long dp[n + 2];
  for (int i = 0; i < (int)n; i++) {
    if (i == 0) {
      dp[i] = (str[i] == '-' ? -1 : 1);
      continue;
    }
    dp[i] = dp[i - 1] + (str[i] == '-' ? -1 : 1);
  }
  long long int ans = 0;
  long long int val = 0;
  for (int i = 0; i < (int)n; i++) {
    if (val + dp[i] < 0) {
      long long t = i + 1;
      ans += min(t, n);
      long long neon = -1;
      val = neon * dp[i];
    }
  }
  ans += (n);
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  if (tt) {
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  } else {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
string str;
const int tt = 1;
void solve() {
  long long int m, n, a, b, c;
  cin >> str;
  n = str.size();
  long long dp[n + 2];
  for (int i = 0; i < (int)n; i++) {
    if (i == 0) {
      dp[i] = (str[i] == '-' ? -1 : 1);
      continue;
    }
    dp[i] = dp[i - 1] + (str[i] == '-' ? -1 : 1);
  }
  long long int ans = 0;
  long long int val = 0;
  for (int i = 0; i < (int)n; i++) {
    if (val + dp[i] < 0) {
      long long t = i + 1;
      ans += min(t, n);
      long long neon = -1;
      val = neon * dp[i];
    }
  }
  ans += (n);
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  if (tt) {
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  } else {
    solve();
  }
  return 0;
}

```