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.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;

/**
 * User :- sudhakar
 * Date :- 29/05/20
 * Time :- 9:56 AM
 */

public class NewTheatreSquare {
    static final int MOD = 1000000007;

    /*Inner class for fast input*/
    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Reader() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String readLine() throws IOException {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

        public String nextString() throws IOException {
            byte c = read();
            while (Character.isWhitespace(c)) {
                c = read();
            }
            StringBuilder builder = new StringBuilder();
            builder.append((char) c);
            c = read();
            while (!Character.isWhitespace(c)) {
                builder.append((char) c);
                c = read();
            }
            return builder.toString();
        }

        public char nextChar() throws IOException {
            byte c = read();
            while (Character.isWhitespace(c)) {
                c = read();
            }
            return (char) c;
        }

        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        public int[] nextIntArray(Reader r, int size) throws IOException {
            int[] arr = new int[size + 1];
            for (int i = 1; i <= size; i++)
                arr[i] = r.nextInt();
            return arr;
        }

        public long[] nextLongArray(Reader r, int size) throws IOException {
            long[] arr = new long[size + 1];
            for (int i = 1; i <= size; i++)
                arr[i] = r.nextLong();
            return arr;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            if (din == null)
                return;
            din.close();
        }
    }

    private static double min(double a, double b) {
        return Math.min(a, b);
    }

    private static double max(double a, double b) {
        return Math.max(a, b);
    }


    public static void main(String[] args) throws IOException {
        Reader r = new Reader();
        int T = r.nextInt();

        while (T-- > 0) {
            int n = r.nextInt();
            int m = r.nextInt();
            int x = r.nextInt();
            int y = r.nextInt();

            char[][] sqr = new char[n + 1][m + 1];
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= m; j++) {
                    sqr[i][j] = r.nextChar();
                }
            }

            long sum = 0;
            for (int i = 1; i <= n; i++) {
                int count = 0;
                for (int j = 1; j <= m; j++) {
                    while (j <= m && sqr[i][j] == '.') {
                        count++;
                        j++;
                    }
                    int p = (int) min(2 * x, y);
                    sum += p * (count / 2) + x * (count % 2);
                    count = 0;
                }
            }
            System.out.println(sum);

        }
    }
}
