description: There is a chessboard of size n by n. The square in the i-th row from top and j-th column from the left is labelled (i,j).
  
  Currently, Gregor has some pawns in the n-th row. There are also enemy pawns in the 1-st row. On one turn, Gregor moves one of his pawns. A pawn can move one square up (from (i,j) to (i-1,j)) if there is no pawn in the destination square. Additionally, a pawn can move one square diagonally up (from (i,j) to either (i-1,j-1) or (i-1,j+1)) if and only if there is an enemy pawn in that square. The enemy pawn is also removed.
  
  Gregor wants to know what is the maximum number of his pawns that can reach row 1?
  
  Note that only Gregor takes turns in this game, and the enemy pawns never move. Also, when Gregor's pawn reaches row 1, it is stuck and cannot make any further moves.
  
  Input
  
  The first line of the input contains one integer t (1≤ t≤ 2⋅ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of three lines. The first line contains a single integer n (2≤ n≤ 2⋅{10}^{5}) — the size of the chessboard.
  
  The second line consists of a string of binary digits of length n, where a 1 in the i-th position corresponds to an enemy pawn in the i-th cell from the left, and 0 corresponds to an empty cell.
  
  The third line consists of a string of binary digits of length n, where a 1 in the i-th position corresponds to a Gregor's pawn in the i-th cell from the left, and 0 corresponds to an empty cell.
  
  It is guaranteed that the sum of n across all test cases is less than 2⋅{10}^{5}.
  
  Output
  
  For each test case, print one integer: the maximum number of Gregor's pawns which can reach the 1-st row.
  
  Example
  
  Input
  
  
  4
  3
  000
  111
  4
  1111
  1111
  3
  010
  010
  5
  11001
  00000
  
  
  Output
  
  
  3
  4
  0
  0
  
  Note
  
  In the first example, Gregor can simply advance all 3 of his pawns forward. Thus, the answer is 3.
  
  In the second example, Gregor can guarantee that all 4 of his pawns reach the enemy row, by following the colored paths as demonstrated in the diagram below. Remember, only Gregor takes turns in this "game"!
  
  <image>
  
  In the third example, Gregor's only pawn is stuck behind the enemy pawn, and cannot reach the end.
  
  In the fourth example, Gregor has no pawns, so the answer is clearly 0.
solution: //Utilities
  import java.io.*;
  import java.util.*;
  
  public class B {
  	static int t;
  	static int n;
  	static char[] enemy, a;
  	static boolean[] done;
  	static int res;
  	
  	public static void main(String[] args) throws IOException {
  		t = in.iscan();
  		while (t-- > 0) {
  			n = in.iscan();
  			enemy = in.sscan().toCharArray();
  			a = in.sscan().toCharArray();
  			done = new boolean[n];
  			res = 0;
  			for (int i = 0; i < n; i++) {
  				if (a[i] == '1') {
  					if (enemy[i] == '1') {
  						if (i - 1 >= 0 && enemy[i-1] == '1' && !done[i-1]) {
  							done[i-1] = true;
  							res++;
  						}
  						else if (i + 1 < n && enemy[i+1] == '1' && !done[i+1]) {
  							done[i+1] = true;
  							res++;
  						}
  					}
  					else {
  						res++;
  					}
  				}
  			}
  			out.println(res);
  		}
  		out.close();
  	} 
  
  	static INPUT in = new INPUT(System.in);
  	static PrintWriter out = new PrintWriter(System.out);
  	private static class INPUT {
  
  		private InputStream stream;
  		private byte[] buf = new byte[1024];
  		private int curChar, numChars;
  
  		public INPUT (InputStream stream) {
  			this.stream = stream;
  		}
  
  		public INPUT (String file) throws IOException {
  			this.stream = new FileInputStream (file);
  		}
  
  		public int cscan () throws IOException {
  			if (curChar >= numChars) {
  				curChar = 0;
  				numChars = stream.read (buf);
  			}
  			
  			if (numChars == -1)
  				return numChars;
  
  			return buf[curChar++];
  		}
  
  		public int iscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			int res = 0;
  
  			do {
  				res = (res << 1) + (res << 3);
  				res += c - '0';
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res * sgn;
  		}
  
  		public String sscan () throws IOException {
  			int c = cscan ();
  			
  			while (space (c))
  				c = cscan ();
  
  			StringBuilder res = new StringBuilder ();
  
  			do {
  				res.appendCodePoint (c);
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res.toString ();
  		}
  
  		public double dscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			double res = 0;
  
  			while (!space (c) && c != '.') {
  				if (c == 'e' || c == 'E')
  					return res * UTILITIES.fast_pow (10, iscan ());
  				
  				res *= 10;
  				res += c - '0';
  				c = cscan ();
  			}
  
  			if (c == '.') {
  				c = cscan ();
  				double m = 1;
  
  				while (!space (c)) {
  					if (c == 'e' || c == 'E')
  						return res * UTILITIES.fast_pow (10, iscan ());
  
  					m /= 10;
  					res += (c - '0') * m;
  					c = cscan ();
  				}
  			}
  
  			return res * sgn;
  		}
  
  		public long lscan () throws IOException {
  			int c = cscan (), sgn = 1;
  			
  			while (space (c))
  				c = cscan ();
  
  			if (c == '-') {
  				sgn = -1;
  				c = cscan ();
  			}
  
  			long res = 0;
  
  			do {
  				res = (res << 1) + (res << 3);
  				res += c - '0';
  				c = cscan ();
  			}
  			while (!space (c));
  
  			return res * sgn;
  		}
  
  		public boolean space (int c) {
  			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
  		}
  	}
  
  	public static class UTILITIES {
  
  		static final double EPS = 10e-6;
  		
  		public static void sort(int[] a, boolean increasing) {
  			ArrayList<Integer> arr = new ArrayList<Integer>();
  			int n = a.length;
  			for (int i = 0; i < n; i++) {
  				arr.add(a[i]);
  			}
  			Collections.sort(arr);
  			for (int i = 0; i < n; i++) {
  				if (increasing) {
  					a[i] = arr.get(i);
  				}
  				else {
  					a[i] = arr.get(n-1-i);
  				}
  			}
  		}
  		
  		public static void sort(long[] a, boolean increasing) {
  			ArrayList<Long> arr = new ArrayList<Long>();
  			int n = a.length;
  			for (int i = 0; i < n; i++) {
  				arr.add(a[i]);
  			}
  			Collections.sort(arr);
  			for (int i = 0; i < n; i++) {
  				if (increasing) {
  					a[i] = arr.get(i);
  				}
  				else {
  					a[i] = arr.get(n-1-i);
  				}
  			}
  		}
  		
  		public static void sort(double[] a, boolean increasing) {
  			ArrayList<Double> arr = new ArrayList<Double>();
  			int n = a.length;
  			for (int i = 0; i < n; i++) {
  				arr.add(a[i]);
  			}
  			Collections.sort(arr);
  			for (int i = 0; i < n; i++) {
  				if (increasing) {
  					a[i] = arr.get(i);
  				}
  				else {
  					a[i] = arr.get(n-1-i);
  				}
  			}
  		}
  
  		public static int lower_bound (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;
  		}
  
  		public static int upper_bound (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;
  		}
  
  		public static long gcd (long a, long b) {
  			return b == 0 ? a : gcd (b, a % b);
  		}
  
  		public static long lcm (long a, long b) {
  			return a * b / gcd (a, b);
  		}
  
  		public static long fast_pow_mod (long b, long x, int mod) {
  			if (x == 0) return 1;
  			if (x == 1) return b;
  			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;
  
  			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
  		}
  
  		public static int fast_pow (int b, int x) {
  			if (x == 0) return 1;
  			if (x == 1) return b;
  			if (x % 2 == 0) return fast_pow (b * b, x / 2);
  
  			return b * fast_pow (b * b, x / 2);
  		}
  
  		public static long choose (long n, long k) {
  			k = Math.min (k, n - k);
  			long val = 1;
  
  			for (int i = 0; i < k; ++i)
  				val = val * (n - i) / (i + 1);
  
  			return val;
  		}
  
  		public static long permute (int n, int k) {
  			if (n < k) return 0;
  			long val = 1;
  
  			for (int i = 0; i < k; ++i)
  				val = (val * (n - i));
  
  			return val;
  		}
  		
  		// start of permutation and lower/upper bound template
  		public static void nextPermutation(int[] nums) {
  		    //find first decreasing digit
  		    int mark = -1;
  		    for (int i = nums.length - 1; i > 0; i--) {
  		        if (nums[i] > nums[i - 1]) {
  		            mark = i - 1;
  		            break;
  		        }
  		    }
  		 
  		    if (mark == -1) {
  		        reverse(nums, 0, nums.length - 1);
  		        return;
  		    }
  		 
  		    int idx = nums.length-1;
  		    for (int i = nums.length-1; i >= mark+1; i--) {
  		        if (nums[i] > nums[mark]) {
  		            idx = i;
  		            break;
  		        }
  		    }
  		 
  		    swap(nums, mark, idx);
  		 
  		    reverse(nums, mark + 1, nums.length - 1);
  		}
  		 
  		public static void swap(int[] nums, int i, int j) {
  		    int t = nums[i];
  		    nums[i] = nums[j];
  		    nums[j] = t;
  		}
  		 
  		public static void reverse(int[] nums, int i, int j) {
  		    while (i < j) {
  		        swap(nums, i, j);
  		        i++;
  		        j--;
  		    }
  		}
  		
  		static int lower_bound (int[] arr, int hi, int cmp) {
  			int low = 0, high = hi, mid = -1;
  			while (low < high) {
  				mid = (low + high) / 2;
  				if (arr[mid] >= cmp) high = mid;
  				else low = mid + 1;
  			}
  			return low;
  		}
  	 
  		static int upper_bound (int[] arr, int hi, int cmp) {
  			int low = 0, high = hi, mid = -1;
  			while (low < high) {
  				mid = (low + high) / 2;
  				if (arr[mid] > cmp) high = mid;
  				else low = mid + 1;
  			}
  			return low;
  		}
  		// end of permutation and lower/upper bound template
  	}
  }
  