You might have remembered Theatre square from the [problem 1A](https://codeforces.com/problemset/problem/1/A). Now it's finally getting repaved.

The square still has a rectangular shape of n × m meters. However, the picture is about to get more complicated now. Let a_{i,j} be the j-th square in the i-th row of the pavement.

You are given the picture of the squares:

  * if a_{i,j} =  "*", then the j-th square in the i-th row should be black; 
  * if a_{i,j} =  ".", then the j-th square in the i-th row should be white. 



The black squares are paved already. You have to pave the white squares. There are two options for pavement tiles:

  * 1 × 1 tiles — each tile costs x burles and covers exactly 1 square; 
  * 1 × 2 tiles — each tile costs y burles and covers exactly 2 adjacent squares of the same row. Note that you are not allowed to rotate these tiles or cut them into 1 × 1 tiles.



You should cover all the white squares, no two tiles should overlap and no black squares should be covered by tiles.

What is the smallest total price of the tiles needed to cover all the white squares?

Input

The first line contains a single integer t (1 ≤ t ≤ 500) — the number of testcases. Then the description of t testcases follow.

The first line of each testcase contains four integers n, m, x and y (1 ≤ n ≤ 100; 1 ≤ m ≤ 1000; 1 ≤ x, y ≤ 1000) — the size of the Theatre square, the price of the 1 × 1 tile and the price of the 1 × 2 tile.

Each of the next n lines contains m characters. The j-th character in the i-th line is a_{i,j}. If a_{i,j} =  "*", then the j-th square in the i-th row should be black, and if a_{i,j} =  ".", then the j-th square in the i-th row should be white.

It's guaranteed that the sum of n × m over all testcases doesn't exceed 10^5.

Output

For each testcase print a single integer — the smallest total price of the tiles needed to cover all the white squares in burles.

Example

Input


4
1 1 10 1
.
1 2 10 1
..
2 1 10 1
.
.
3 3 3 7
..*
*..
.*.


Output


10
1
20
18

Note

In the first testcase you are required to use a single 1 × 1 tile, even though 1 × 2 tile is cheaper. So the total price is 10 burles.

In the second testcase you can either use two 1 × 1 tiles and spend 20 burles or use a single 1 × 2 tile and spend 1 burle. The second option is cheaper, thus the answer is 1.

The third testcase shows that you can't rotate 1 × 2 tiles. You still have to use two 1 × 1 tiles for the total price of 20.

In the fourth testcase the cheapest way is to use 1 × 1 tiles everywhere. The total cost is 6 ⋅ 3 = 18.
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class B {
	static InputReader scn = new InputReader(System.in);
	static OutputWriter out = new OutputWriter(System.out);

	public static void main(String[] args) {
		int t = scn.nextInt();
		while (t-- > 0) {
			int n = scn.nextInt();
			int m = scn.nextInt();
			int x = scn.nextInt();
			int y = scn.nextInt();
			System.out.println();
			int ans = 0;
			int l = 0;
			while (l < n) {
				char[] ch = new char[m];
				String s = scn.next();
				for (int i = 0; i < m; i++) {
					ch[i] = s.charAt(i);
				}
				if (m == 1 && ch[0] == '.')
					ans += x;
				for (int i = 0; i < m - 1; i++) {
					char ch1 = ch[i];
					char ch2 = ch[i + 1];
					if (ch1 == ch2 && ch1 == '.') {
						ans = ans + Math.min(2 * x, y);
						i++;
					} else {
						if (ch1 == '.') {
							ans = ans + x;
						}
					}
				}
				if (m >= 2 && ch[m - 1] == '.' && ch[m - 1] != ch[m - 2]) {
					ans = ans + x;
				}
//				System.out.println(l + " " + ans);
				l++;
			}
			out.println(ans);
		}
		out.close();
	}

	public static HashMap<Integer, Integer> CountFrequencies(int[] arr) {
		HashMap<Integer, Integer> map = new HashMap<>();
		for (int i = 0; i < arr.length; i++) {
			int a = arr[i];
			if (map.containsKey(a)) {
				map.put(a, map.get(a) + 1);
			} else {
				map.put(a, 1);
			}
		}
		return map;
	}

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

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

	public static ArrayList<String> GetSubSequences(String s) {
		if (s.length() == 0) {
			ArrayList<String> br = new ArrayList<>();
			br.add("");
			return br;
		}
		char ch = s.charAt(0);
		String ms = s.substring(1);

		ArrayList<String> rr = GetSubSequences(ms);

		ArrayList<String> mr = new ArrayList<>();

		int t = rr.size();
		for (int i = 0; i < t; i++) {
			mr.add(rr.get(i));
			mr.add(ch + rr.get(i));

		}

		return mr;

	}

	public static void printSubsequences(String[] str, int n) {
		/* Number of subsequences is (2**n -1) */
		int opsize = (int) Math.pow(2, n);
		for (int counter = 1; counter < opsize; counter++) {
			for (int j = 0; j < n; j++) {
				/*
				 * Check if jth bit in the counter is set If set then print jth element from
				 * arr[]
				 */

				if (BigInteger.valueOf(counter).testBit(j))
					System.out.print(str[j] + " ");
			}
			System.out.println();
		}
	}

	public static int[] readArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = scn.nextInt();
		}
		return a;
	}

	static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a) {
			l.add(i);
		}
		Collections.sort(l);
		for (int i = 0; i < a.length; i++) {
			a[i] = l.get(i);
		}
	}

	public static int binarySearch(int arr[], int l, int r, int x) {
		if (r >= l) {
			int mid = l + (r - l) / 2;
			if (arr[mid] == x)
				return mid;
			if (arr[mid] > x)
				return binarySearch(arr, l, mid - 1, x);
			return binarySearch(arr, mid + 1, r, x);
		}
		return -1;
	}

	public static int firstOccurence(int array1[], int low, int high, int x, int n) {
		if (low <= high) {
			int mid = low + (high - low) / 2;
			if ((mid == 0 || x > array1[mid - 1]) && array1[mid] == x)
				return mid;
			else if (x > array1[mid])
				return firstOccurence(array1, (mid + 1), high, x, n);
			else
				return firstOccurence(array1, low, (mid - 1), x, n);
		}
		return -1;
	}

	public static int lastOccurence(int array2[], int low, int high, int x, int n) {
		if (low <= high) {
			int mid = low + (high - low) / 2;
			if ((mid == n - 1 || x < array2[mid + 1]) && array2[mid] == x)
				return mid;
			else if (x < array2[mid])
				return lastOccurence(array2, low, (mid - 1), x, n);
			else
				return lastOccurence(array2, (mid + 1), high, x, n);
		}
		return -1;
	}

	public static boolean isPrime(long n) {
		if (n < 2)
			return false;
		if (n == 2 || n == 3)
			return true;
		if (n % 2 == 0 || n % 3 == 0)
			return false;
		long sqrtN = (int) Math.sqrt(n) + 1;
		for (long i = 6L; i <= sqrtN; i += 6) {
			if (n % (i - 1) == 0 || n % (i + 1) == 0)
				return false;
		}
		return true;
	}

	public static void quickSort(int[] arr, int lo, int hi) {
		if (lo >= hi) {
			return;
		}
		int mid = (lo + hi) / 2;
		int pivot = arr[mid];
		int left = lo;
		int right = hi;
		while (left <= right) {
			while (arr[left] < pivot) {
				left++;
			}
			while (arr[right] > pivot) {
				right--;
			}
			if (left <= right) {
				int temp = arr[left];
				arr[left] = arr[right];
				arr[right] = temp;
				left++;
				right--;
			}
		}
		quickSort(arr, lo, right);
		quickSort(arr, left, hi);
	}

	static class InputReader {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;

		public InputReader(InputStream stream) {
			this.stream = stream;
		}

		public int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public boolean isSpaceChar(int c) {
			if (filter != null)
				return filter.isSpaceChar(c);
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public String next() {
			return nextLine();
		}

		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}

	static class OutputWriter {
		private final PrintWriter writer;

		public OutputWriter(OutputStream outputStream) {
			writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
		}

		public OutputWriter(Writer writer) {
			this.writer = new PrintWriter(writer);
		}

		public void print(Object... objects) {
			for (int i = 0; i < objects.length; i++) {
				if (i != 0) {
					writer.print(' ');
//                    writer.print(1);
				}
				writer.print(objects[i]);
			}
		}

		public void println(Object... objects) {
			print(objects);
			writer.println();
		}

		public void close() {
			writer.close();
		}
	}

}
