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.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class C {
	private static PrintWriter out = new PrintWriter(System.out);

	public static void solve() {
		In in = new In();
		int tests = in.ni();
		while (tests-- > 0) {
			int n = in.ni();
			char[] s = in.ns().toCharArray();
			char[] t = in.ns().toCharArray();
			long ans = 0;
			for (int i = 0; i < n; i++) {
				if (s[i] != t[i]) {
					ans += 2;
				} else if (s[i] == '0') {
					if (i + 1 < n) {
						if (s[i + 1] != s[i] && t[i + 1] != s[i]) {
							ans += 2;
							i++;
						} else {
							ans++;
						}
					} else {
						ans++;
					}
				} else { // s[i] == t[i] == '1'
					if (i + 1 < n && (s[i + 1] != s[i] || t[i + 1] != s[i])) {
						ans += 2;
						i++;
					}
				}
			}
			out.println(ans);

		}
	}

	public static void main(String[] args) throws IOException {
		// long start = System.nanoTime();
		solve();
		// System.out.println("Elapsed: " + (System.nanoTime() - start) / 1000000 +
		// "ns");
		out.flush();
	}

	@SuppressWarnings("unused")
	private static class In {
		final private static int BUFFER_SIZE = 1024;
		private byte[] buf;
		private InputStream is;
		private int buflen;
		private int bufptr;

		public In() {
			is = System.in;
			buf = new byte[BUFFER_SIZE];
			buflen = bufptr = 0;
		}

		public In(String input) {
			is = new ByteArrayInputStream(input.getBytes());
			buf = new byte[BUFFER_SIZE];
			buflen = bufptr = 0;
		}

		public int readByte() {
			if (bufptr >= buflen) {
				try {
					buflen = is.read(buf);
				} catch (IOException ioe) {
					throw new InputMismatchException();
				}
				bufptr = 0;
			}
			if (buflen <= 0)
				return -1;
			return buf[bufptr++];
		}

		public boolean isSpaceChar(int c) {
			return !(c >= 33 && c <= 126);
		}

		public int skip() {
			int b;
			while ((b = readByte()) != -1 && isSpaceChar(b))
				;
			return b;
		}

		/* Next character */
		public char nc() {
			return (char) skip();
		}

		/* Next double */
		public double nd() {
			return Double.parseDouble(ns());
		}

		/* Next string */
		public String ns() {
			final StringBuilder sb = new StringBuilder();
			int b = skip();
			while (!isSpaceChar(b)) {
				sb.appendCodePoint(b);
				b = readByte();
			}
			return sb.toString();
		}

		/* Next integer */
		public int ni() {
			boolean minus = false;
			int num = 0;
			int b;
			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
				;
			if (b == '-') {
				minus = true;
				b = readByte();
			}
			while (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
				b = readByte();
			}
			return minus ? -num : num;
		}

		/* Next long */
		public long nl() {
			boolean minus = false;
			long num = 0;
			int b;
			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
				;
			if (b == '-') {
				minus = true;
				b = readByte();
			}
			while (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
				b = readByte();
			}
			return minus ? -num : num;
		}

		/* Next integer 1D array */
		public Integer[] nia(int n) {
			final Integer[] arr = new Integer[n];
			for (int i = 0; i < n; i++)
				arr[i] = ni();
			return arr;
		}

		/* Next long 1D array */
		public long[] nla(int n) {
			final long[] arr = new long[n];
			for (int i = 0; i < n; i++)
				arr[i] = nl();
			return arr;
		}

		/* Next string 1D array */
		public String[] nsa(int n) {
			final String[] arr = new String[n];
			for (int i = 0; i < n; i++)
				arr[i] = ns();
			return arr;
		}

		/* Next char 1D array */
		public char[] nca(int n) {
			final char[] arr = new char[n];
			for (int i = 0; i < n; i++)
				arr[i] = nc();
			return arr;
		}

		/* Next double 1D array */
		public double[] nda(int n) {
			final double[] arr = new double[n];
			for (int i = 0; i < n; i++)
				arr[i] = nc();
			return arr;
		}

		/* Next integer matrix */
		public int[][] nim(int n, int m) {
			final int[][] arr = new int[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					arr[i][j] = ni();
			return arr;
		}

		/* Next long matrix */
		public long[][] nlm(int n, int m) {
			final long[][] arr = new long[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					arr[i][j] = nl();
			return arr;
		}

		/* Next string matrix */
		public String[][] nsm(int n, int m) {
			final String[][] arr = new String[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					arr[i][j] = ns();
			return arr;
		}

		/* Next char matrix */
		public char[][] ncm(int n, int m) {
			final char[][] arr = new char[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					arr[i][j] = nc();
			return arr;
		}

		/* Next double matrix */
		public double[][] ndm(int n, int m) {
			final double[][] arr = new double[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					arr[i][j] = nd();
			return arr;
		}

		public static void log(Object... o) {
			System.out.println(Arrays.deepToString(o));
		}
	}

}
