Educational Codeforces Round 90 (Rated for Div. 2) C. Pluses and Minuses

Educational Codeforces Round 90 (Rated for Div. 2) C. Pluses and Minuses

题目链接
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|≤1e6) consisting only of characters + and -.

It’s guaranteed that sum of |s| over all test cases doesn’t exceed 1e6.

Output

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

Example

input

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

output

7
9
6

阅读理解题，维护一个前缀即可，每出现一次小于0就加到答案里，最后加上字符串长度即可，注意会爆 int，AC代码如下：

#include
using namespace std;
typedef long long ll;
int main()
{
     
    int t;
    cin>>t;
    while(t--) {
     
        string s;
        cin>>s;
        ll i,pre=0,ans=0;
        for(i=0;i<s.size();i++){
     
            pre+=(s[i]=='+'?1:-1);
            if(pre<0) {
     ans+=i+1;pre++;}
        }
        ans+=s.size();
        cout<<ans<<endl;
    }
    return 0;
}