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.
import java.util.*;
import java.io.*;
public class Main {
	public static void main(String[] args) throws IOException, InterruptedException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int jj = sc.nextInt();
		while(jj-->0) {
			int n = sc.nextInt();
			String s = sc.nextLine();
			String v = sc.nextLine();
			
			int ans = 0;
			for(int i = 0;i<s.length();i++) {
				if(s.charAt(i)=='0'&&v.charAt(i)=='1') {
					ans+=2;
				}
				if(s.charAt(i)=='1'&&v.charAt(i)=='0') {
					ans+=2;
				}
				if(s.charAt(i)=='0'&&v.charAt(i)=='0') {
					if(i==s.length()-1) {
						ans+=1;
					}
					else {
						if(s.charAt(i+1)=='1'&&v.charAt(i+1)=='1') {
							ans+=2;
							i++;
						}
						else {
							ans += 1;
						}
					}
				}
				if(s.charAt(i)=='1'&&v.charAt(i)=='1') {
					while(i<s.length()) {
						if(s.charAt(i)!='1'||v.charAt(i)!='1') {
							ans+=2;
							
							break;
						}
						i++;
					}
				}
			}
			pw.println(ans);
		}
				
		
		pw.flush();
	}
	
	
}

class edge implements Comparable<edge> {
	int from;
	int dest;
	double cost;

	public edge(int from, int dest, double cost) {
		this.from = from;
		this.dest = dest;
		this.cost = cost;

	}
public String toString() {
	return from+" "+dest+" "+cost;
}
	@Override
	public int compareTo(edge o) {
//		return cost - o.cost;
		if (cost < o.cost)
			return -1;
		if (cost > o.cost)
			return 1;
		return 0;
	}

}

class pair {
	int a;
	int b;

	public pair(int a, int b) {
		this.a = a;
		this.b = b;
	}

}

class Scanner {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s) {
		br = new BufferedReader(new InputStreamReader(s));
	}

	public String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	public long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	public String nextLine() throws IOException {
		return br.readLine();
	}

	public double nextDouble() throws IOException {
		String x = next();
		StringBuilder sb = new StringBuilder("0");
		double res = 0, f = 1;
		boolean dec = false, neg = false;
		int start = 0;
		if (x.charAt(0) == '-') {
			neg = true;
			start++;
		}
		for (int i = start; i < x.length(); i++)
			if (x.charAt(i) == '.') {
				res = Long.parseLong(sb.toString());
				sb = new StringBuilder("0");
				dec = true;
			} else {
				sb.append(x.charAt(i));
				if (dec)
					f *= 10;
			}
		res += Long.parseLong(sb.toString()) / f;
		return res * (neg ? -1 : 1);
	}

	public boolean ready() throws IOException {
		return br.ready();
	}
}