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.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class C {
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int t = sc.nextInt();
		for (int i = 0; i < t; i++) {
			int n = sc.nextInt();
			int[][] x = new int[2][n];
			int[] cSum = new int[n];
			char[] s1 = sc.nextToken().toCharArray();
			char[] s2 = sc.nextToken().toCharArray();
			for (int j = 0; j < n; j++) {
				if (s1[j] == '0') {
					x[0][j] = 0;
				} else {
					x[0][j] = 1;
				}
				if (s2[j] == '0') {
					x[1][j] = 0;
				} else {
					x[1][j] = 1;
				}
				cSum[j] = x[0][j] + x[1][j];
			}

			int cc = 0;
			int ans = 0;
			while (cc < n) {
				if (cc == n - 1) {
					if (cSum[cc] == 0) {
						ans += 1;
						cc += 1;
					} else if (cSum[cc] == 1) {
						ans += 2;
						cc += 1;
					} else {
						ans += 0;
						cc += 1;
					}
				} else {

					if (cSum[cc] == 0) { // 00
						if (cSum[cc + 1] == 2) {
							ans += 2;
							cc += 2;
						} else {
							ans += 1;
							cc += 1;
						}
					} else if (cSum[cc] == 1) {
						ans += 2;
						cc += 1;
					} else {
						if (cSum[cc + 1] == 0) {
							ans += 2;
							cc += 2;
						} else {
							ans += 0;
							cc += 1;
						}
					}
				}

			}
			System.out.println(ans);
		}

	}

	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
		}

		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String nextToken() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					// TODO Auto-generated catch block
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(nextToken());
		}

		long nextLong() {
			return Long.parseLong(nextToken());
		}

		double nextDouble() {
			return Double.parseDouble(nextToken());
		}
	}

}
