Problem: You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'.

A string is called a regular bracket sequence (RBS) if it's of one of the following types: 

  * empty string; 
  * '(' + RBS + ')'; 
  * '[' + RBS + ']'; 
  * RBS + RBS. 



where plus is a concatenation of two strings.

In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order.

What is the maximum number of moves you can perform?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2 ⋅ 10^5.

Output

For each testcase print a single integer — the maximum number of moves you can perform on a given string s.

Example

Input


5
()
[]()
([)]
)]([
)[(]


Output


1
2
2
0
1

Note

In the first example you can just erase the whole string.

In the second example you can first erase the brackets on positions 1 and 2: "[]()", then "()" is left. After that you can erase it whole. You could erase the whole string from the beginning but you would get one move instead of two.

In the third example you can first erase the brackets on positions 1 and 3: "([)]". They form an RBS "()". Then "[]" is left, so you can erase it whole.

In the fourth example there is no subsequence that is an RBS, so you can't perform a move at all.

In the fifth example you can erase the brackets on positions 2 and 4: ")[(]" and get ")(" as a result. You can erase nothing from it.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define lli long long int
#define fop(i,m,n) for (int i=(m);i<(n);i++)
#define FOP(i,m,n) for (int i=(m)-1;i>=(n);--i)
#define test(x) cout << #x << ' ' << x << endl;
#define printv(a) {\
    for (auto i : a) cout << i << ' ';\
    cout << '\n';\
}
#define eb emplace_back
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
#define pll pair<lli,lli>
#define ld long double;
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define X first
#define Y second
auto SEED = chrono::steady_clock::now().time_since_epoch().count();
mt19937 rng(SEED);
const int mod = 1e9 + 7, x = 864197532, N = 100087, logN = 18, K = 500, C = 100;

void solve() {
    string s;
    cin >> s;
    int n = s.length();
    int ans = 0;
    int a = 0, b = 0;
    fop (i,0,n) {
        if (s[i] == '(') a++;
        else if (s[i] == '[') b++;
        else if (s[i] == ')' && a) ans++, a--;
        else if (s[i] == ']' && b) ans++, b--;
    }
    cout << ans << endl;
}

int main () {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}