A binary string is a string that consists of characters 0 and 1. A bi-table is a table that has exactly two rows of equal length, each being a binary string.

Let \operatorname{MEX} of a bi-table be the smallest digit among 0, 1, or 2 that does not occur in the bi-table. For example, \operatorname{MEX} for \begin{bmatrix} 0011\\\ 1010 \end{bmatrix} is 2, because 0 and 1 occur in the bi-table at least once. \operatorname{MEX} for \begin{bmatrix} 111\\\ 111 \end{bmatrix} is 0, because 0 and 2 do not occur in the bi-table, and 0 < 2.

You are given a bi-table with n columns. You should cut it into any number of bi-tables (each consisting of consecutive columns) so that each column is in exactly one bi-table. It is possible to cut the bi-table into a single bi-table — the whole bi-table.

What is the maximal sum of \operatorname{MEX} of all resulting bi-tables can be?

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.

The first line of the description of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of columns in the bi-table.

Each of the next two lines contains a binary string of length n — the rows of the bi-table.

It's guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case print a single integer — the maximal sum of \operatorname{MEX} of all bi-tables that it is possible to get by cutting the given bi-table optimally.

Example

Input


4
7
0101000
1101100
5
01100
10101
2
01
01
6
000000
111111


Output


8
8
2
12

Note

In the first test case you can cut the bi-table as follows:

  * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
  * \begin{bmatrix} 10\\\ 10 \end{bmatrix}, its \operatorname{MEX} is 2.
  * \begin{bmatrix} 1\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 0.
  * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
  * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.
  * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.



The sum of \operatorname{MEX} is 8.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define f first
#define s second
#define all(vec) begin(vec), end(vec)
#define pf push_front
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define mk make_pair
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ll,int> pli;

int mex(set<int>& st){
    for(int i = 0; i <= 2; i++){
        if(!st.count(i)){
            return i;
        }
    }
    assert(false);
}
void solve(){
    int n; cin >> n;
    string s, t;
    cin >> s >> t;
    int ans = 0;
    set<int> st;
    for(int i = 0; i < n; i++){
        st.clear();
        st.insert(s[i]-'0');
        st.insert(t[i]-'0');
        int premex = mex(st);
        if(premex == 2){
            ans += 2;
            continue;
        }
        if(premex == 1){
            if(i == n-1){
                ans += 1;
                break;
            }
            st.insert(s[i+1]-'0');
            st.insert(t[i+1]-'0');
            int nxtmex = mex(st);
            if(nxtmex == 1){
                ans += 1;
            }else{ // == 2
                ans += 2;
                i++;
            }
            continue;
        }
        i++;
        while(i < n){
            st.insert(s[i]-'0');
            st.insert(t[i]-'0');
            if(mex(st) == 2){
                break;
            }
            i++;
        }
        ans += mex(st);
    }
    cout << ans << '\n';
}
int main(){
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
//    freopen("input.in", "r", stdin);
//    freopen("output.out", "w", stdout);
    int tc = 1;
    cin >> tc;
    while(tc--){
        solve();
    }
}
