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.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Random;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.io.*;

	public class A {
		
		BufferedReader br;
		StringTokenizer st;
		
		
		public A(){ // constructor
			
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next(){
			
			while (st == null || !st.hasMoreElements()){
				try{
					st = new StringTokenizer(br.readLine());
				}
				
				catch (IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int nextInt(){
			
			return Integer.parseInt(next());
		}
		
		long nextLong(){
			
			return Long.parseLong(next());
		}
		
		double nextDouble(){
			return Double.parseDouble(next());
		}

		String nextLine(){
			
			String str = "";
			try{
				
				str = br.readLine();
			}
			catch (IOException e){
				
				e.printStackTrace();
			}
			return str;
		}
		
		static void sort(int[] arr){
			
	        int n = arr.length;
	        
	        Random rnd = new Random();
	        
	        for(int i = 0; i < n; ++i){
	        	
	            int tmp = arr[i];
	            int randomPos = i + rnd.nextInt(n - i);
	            arr[i] = arr[randomPos];
	            arr[randomPos] = tmp;
	        }
	        
	        Arrays.sort(arr);
	   }
		
		static class Pair{
			int loc;
			int speed;
			Pair(int loc, int speed){
				this.loc = loc;
				this.speed = speed;
			}
		}
		
		static long gcd_long(long a, long b) {
			if(b == 0) {
				return a;
			}
			return gcd_long(b, a % b);
		}
		
		static int mod = (int)1e9 + 7;
		static long mul_long = 1;
		
	public static void main(String[] args) throws Exception{
		
		A in = new A();
		PrintWriter out = new PrintWriter(System.out);
		
		// hard work will power dedication
		
		// Rippling CNIL Spri GOOG.
		
		int test = in.nextInt();
		while(test-- > 0) {
			
			int n = in.nextInt();
			String s1 = in.next();
			String s2 = in.next();
			
			int ans = 0;
			
			int oo = 0;
			int zz = 0;
			
			for(int i = 0; i < n; i++) {
				char c1 = s1.charAt(i);
				char c2 = s2.charAt(i);
				
				if(c1 == '0' && c2 == '1' || c1 == '1' && c2 == '0') {
					ans += 2;
					oo = 0;
					zz = 0;
				}
				
				else if(c1 == '0' && c2 == '0') {
					oo++;
					if(zz >= 1) {
						ans += 2;
						zz = 0;
						oo = 0;
					}
					else if(oo > 1){
						ans++;
						oo--;
					}
				}
				
				else if(c1 == '1' && c2 == '1') {
					zz++;
					if(oo == 1) {
						ans += 2;
						zz = 0;
						oo = 0;
					}
				}
				
			}
			
			out.println(ans + oo);
		}
		
		out.flush();
		out.close();
	}
}