They say "years are like dominoes, tumbling one after the other". But would a year fit into a grid? I don't think so.

Limak is a little polar bear who loves to play. He has recently got a rectangular grid with h rows and w columns. Each cell is a square, either empty (denoted by '.') or forbidden (denoted by '#'). Rows are numbered 1 through h from top to bottom. Columns are numbered 1 through w from left to right.

Also, Limak has a single domino. He wants to put it somewhere in a grid. A domino will occupy exactly two adjacent cells, located either in one row or in one column. Both adjacent cells must be empty and must be inside a grid.

Limak needs more fun and thus he is going to consider some queries. In each query he chooses some rectangle and wonders, how many way are there to put a single domino inside of the chosen rectangle?

Input

The first line of the input contains two integers h and w (1 ≤ h, w ≤ 500) – the number of rows and the number of columns, respectively.

The next h lines describe a grid. Each line contains a string of the length w. Each character is either '.' or '#' — denoting an empty or forbidden cell, respectively.

The next line contains a single integer q (1 ≤ q ≤ 100 000) — the number of queries.

Each of the next q lines contains four integers r1i, c1i, r2i, c2i (1 ≤ r1i ≤ r2i ≤ h, 1 ≤ c1i ≤ c2i ≤ w) — the i-th query. Numbers r1i and c1i denote the row and the column (respectively) of the upper left cell of the rectangle. Numbers r2i and c2i denote the row and the column (respectively) of the bottom right cell of the rectangle.

Output

Print q integers, i-th should be equal to the number of ways to put a single domino inside the i-th rectangle.

Examples

Input

5 8
....#..#
.#......
##.#....
##..#.##
........
4
1 1 2 3
4 1 4 1
1 2 4 5
2 5 5 8


Output

4
0
10
15


Input

7 39
.......................................
.###..###..#..###.....###..###..#..###.
...#..#.#..#..#.........#..#.#..#..#...
.###..#.#..#..###.....###..#.#..#..###.
.#....#.#..#....#.....#....#.#..#..#.#.
.###..###..#..###.....###..###..#..###.
.......................................
6
1 1 3 20
2 10 6 30
2 10 7 30
2 2 7 7
1 7 7 7
1 8 7 8


Output

53
89
120
23
0
2

Note

A red frame below corresponds to the first query of the first sample. A domino can be placed in 4 possible ways.

<image>
import java.io.*;
import static java.lang.Math.*;
import java.util.*;
import java.util.function.*;
import java.lang.*;

public class Main {
    final static boolean debug = false;
    final static String fileName = "";
    final static boolean useFiles = false;

    public static void main(String[] args) throws FileNotFoundException {
        long start;
        if (debug)
            start = System.nanoTime();
        InputStream inputStream;
        OutputStream outputStream;
        if (useFiles) {
            inputStream = new FileInputStream(fileName + ".in");
            outputStream = new FileOutputStream(fileName + ".out");
        } else {
            inputStream = System.in;
            outputStream = System.out;
        }
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task(in, out);
        solver.solve();
        if(debug)
            out.println((System.nanoTime() - start) / 1e+9);
        out.close();
    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public double nextDouble(){
        return Double.parseDouble(next());
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public long nextLong(){
        return Long.parseLong(next());
    }

    public byte nextByte(){
        return Byte.parseByte(next());
    }
}

class Task {
    int get(int[][] d, int x1, int x2, int y1, int y2) {
        if (x1 > x2 || y1 > y2) return 0;
        int result = d[x2][y2];
        if (x1 != 0)
            result -= d[x1 - 1][y2];
        if (y1 != 0)
            result -= d[x2][y1 - 1];
        if (x1 != 0 && y1 != 0)
            result += d[x1 - 1][y1 - 1];
        return result;
    }

    public void solve() {
        int n = in.nextInt();
        int m = in.nextInt();
        char[][] a = new char[n][m];
        for (int i = 0; i < n; i++) {
            String s = in.next();
            for (int j = 0; j < m; j++) {
                a[i][j] = s.charAt(j);
            }
        }
        //horiz
        int[][] x = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (i != 0)
                    x[i][j] += x[i - 1][j];
                if (j != 0)
                    x[i][j] += x[i][j - 1];
                if (i != 0 && j != 0)
                    x[i][j] -= x[i - 1][j - 1];
                if (j != 0 && a[i][j] == '.' && a[i][j - 1] == '.')
                    x[i][j]++;
            }
        }
        //ver
        int[][] y = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (i != 0)
                    y[i][j] += y[i - 1][j];
                if (j != 0)
                    y[i][j] += y[i][j - 1];
                if (i != 0 && j != 0)
                    y[i][j] -= y[i - 1][j - 1];
                if (i != 0 && a[i][j] == '.' && a[i - 1][j] == '.')
                    y[i][j]++;
            }
        }
        int q = in.nextInt();
        for (int qq = 0; qq < q; qq++) {
            int x1 = in.nextInt() - 1;
            int y1 = in.nextInt() - 1;
            int x2 = in.nextInt() - 1;
            int y2 = in.nextInt() - 1;
            int result = get(x, x1, x2, y1 + 1, y2) + get(y, x1 + 1, x2, y1, y2);
            out.println(result);
        }
    }

    private InputReader in;
    private PrintWriter out;

    Task(InputReader in, PrintWriter out) {
        this.in = in;
        this.out = out;
    }
}
