#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pll = pair<ll, ll>;
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
template <class T, class U>
void chmin(T& t, const U& u) {
  if (t > u) t = u;
}
template <class T, class U>
void chmax(T& t, const U& u) {
  if (t < u) t = u;
}
template <class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (int i = 0; i < v.size(); i++) {
    if (i) os << ",";
    os << v[i];
  }
  os << "}";
  return os;
}
const ll MOD = 1e9 + 7;
constexpr ll INF = 1e18;
const int MX = 3010;
int t;
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    V<pll> n;
    ll neg = 0;
    for (int i = 0; i < ((int)(s).size()); i++) {
      if (s[i] == '+')
        neg--;
      else
        neg++;
      if (neg > 0 and n.empty())
        n.emplace_back(make_pair(neg, 1ll * (i + 1)));
      else if (neg > 0 and n.back().first < neg)
        n.emplace_back(make_pair(neg, 1ll * (i + 1)));
    }
    ll res = ((int)(s).size());
    if (n.empty()) {
      cout << res << endl;
      continue;
    }
    for (int i = 0; i < ((int)(n).size()); i++) {
      if (i == 0) {
        res += n[i].first * n[i].second;
      } else {
        res += (n[i].first - n[i - 1].first) * n[i].second;
      }
    }
    cout << res << endl;
  }
  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;
using ll = long long;
using pll = pair<ll, ll>;
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
template <class T, class U>
void chmin(T& t, const U& u) {
  if (t > u) t = u;
}
template <class T, class U>
void chmax(T& t, const U& u) {
  if (t < u) t = u;
}
template <class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (int i = 0; i < v.size(); i++) {
    if (i) os << ",";
    os << v[i];
  }
  os << "}";
  return os;
}
const ll MOD = 1e9 + 7;
constexpr ll INF = 1e18;
const int MX = 3010;
int t;
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    V<pll> n;
    ll neg = 0;
    for (int i = 0; i < ((int)(s).size()); i++) {
      if (s[i] == '+')
        neg--;
      else
        neg++;
      if (neg > 0 and n.empty())
        n.emplace_back(make_pair(neg, 1ll * (i + 1)));
      else if (neg > 0 and n.back().first < neg)
        n.emplace_back(make_pair(neg, 1ll * (i + 1)));
    }
    ll res = ((int)(s).size());
    if (n.empty()) {
      cout << res << endl;
      continue;
    }
    for (int i = 0; i < ((int)(n).size()); i++) {
      if (i == 0) {
        res += n[i].first * n[i].second;
      } else {
        res += (n[i].first - n[i - 1].first) * n[i].second;
      }
    }
    cout << res << endl;
  }
  return 0;
}

```