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.io.*;
import java.util.*;

public class Solution {

    /* -------------------------------------- Main Start ----------------------------------- */
    public static void main(String[] args) throws IOException {
        FastReader fs = new FastReader();
        StringBuilder sb = new StringBuilder();
        int t = Integer.parseInt(fs.nextLine().trim());
        while (t-- > 0) {
            int n = fs.nextInt();
            char[] row1 = fs.nextLine().toCharArray();
            char[] row2 = fs.nextLine().toCharArray();
            int temZ = 0, temO = 0;
            int max = 0;
            int prev = -1;
            for (int i = 0; i < n; i++) {
                if (row1[i] == '0' && row2[i] == '0') {
                    temZ++;
                    if (prev == 1) {
                        max += 2;
                        prev = -1;
                    } else {
                        max++;
                        prev = 0;
                    }
                }
                if (row1[i] == '1' && row2[i] == '0') {
                    temO++;
                    temZ++;
                    max += 2;
                    prev = -1;
                }
                if (row1[i] == '0' && row2[i] == '1') {
                    temO++;
                    temZ++;
                    max += 2;
                    prev = -1;
                }
                if (row1[i] == '1' && row2[i] == '1') {
                    temO++;
                    if (prev == 0) {
                        max++;
                        prev = -1;
                    } else {
                        prev = 1;
                    }
                }
            }

            int curr = 2;
            if (temZ == 0) {
                curr = 0;
            } else if (temO == 0) {
                curr = 1;
            }
            System.out.println(Math.max(max, curr));
        }
    }

    /* ---------------------------------------------------------------------------------------------------------------------------------------------------------------------- 
  
      StringBuilder sb = new StringBuilder();
      sb.append(x + "\n");
    
      Collections.sort(arr, (a, b) -> Integer.compare(a[0], b[0]))
      arr.toArray(new int[arr.size()][])
    
     */
    private static final int MOD_1 = 1000000000 + 7;

    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

        public FastReader() throws IOException {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        String nextLine() throws IOException {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        /* --------------------------------------------------------------------------------------------------------------------------------------------------------------------- */
        void printArray(int[] arr) {
            System.out.println(Arrays.toString(arr));
        }

        void printArray(String[] arr) {
            System.out.println(Arrays.toString(arr));
        }

        void printArray(long[] arr) {
            System.out.println(Arrays.toString(arr));
        }

        void print(int data) {
            System.out.println(data);
        }

        void print(String data) {
            System.out.println(data);
        }

        void print(long data) {
            System.out.println(data);
        }

        int[] II(int n) throws IOException {
            int[] d = new int[n];
            String[] arr = nextLine().split(" ");
            for (int i = 0; i < n; i++) {
                d[i] = Integer.parseInt(arr[i]);
            }

            return d;
        }

        String[] IS(int n) throws IOException {
            return nextLine().split(" ");
        }

        long[] IL(int n) throws IOException {
            long[] d = new long[n];
            String[] arr = nextLine().split(" ");
            for (int i = 0; i < n; i++) {
                d[i] = Long.parseLong(arr[i]);
            }
            return d;
        }

        public long gcd(long a, long b) {
            if (a == 0) {
                return b;
            }
            return gcd(b % a, a);
        }

        public long power(long x, long y, long p) {
            long res = 1;
            x = x % p;
            if (x == 0) {
                return 0;
            }

            while (y > 0) {
                if ((y & 1) == 1) {
                    res = (res * x) % p;
                }
                y = y >> 1;
                x = (x * x) % p;
            }
            return res;
        }

        void sieveOfEratosthenes(boolean prime[], int size) {
            Arrays.fill(prime, true);
            prime[0] = prime[1] = false;
            prime[2] = true;

            for (int p = 2; p * p < size; p++) {
                if (prime[p] == true) {
                    for (int i = p * p; i < size; i += p) {
                        prime[i] = false;
                    }
                }
            }
        }

        public long fact(long n) {
            long ans = 1;
            for (int i = 2; i <= n; i++) {
                ans = (ans * i) % MOD_1;
            }
            return ans;
        }

        public long lcm(long a, long b) {
            return (a * b) / gcd(a, b);
        }

        public int log2(int N) {
            int result = (int) (Math.log(N) / Math.log(2));
            return result;
        }

        // find first element in the arr that is equal to the x. 
        // If x is not present in arr then it will return first element that is greater than x
        // It return arr.size() if all element are smaller than x
        public int lowerBound(int[] arr, int x) {
            int low = 0, high = arr.length, mid = -1;

            while (low < high) {
                mid = (low + high) / 2;

                if (arr[mid] >= x) {
                    high = mid;
                } else {
                    low = mid + 1;
                }
            }

            return low;
        }

        // find first element in the arr that is greater than x. 
        // It return arr.size() if all element are smaller than x
        public int upperBound(int[] arr, int x) {
            int low = 0, high = arr.length, mid = -1;

            while (low < high) {
                mid = (low + high) / 2;

                if (arr[mid] > x) {
                    high = mid;
                } else {
                    low = mid + 1;
                }
            }

            return low;
        }

    }

}
