C. New Year and Domino

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 r2iand 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.



解题说明：题目大意是在给定的n,m的矩阵中问在以l1,r1,l2,r2为边界的矩阵1*2的骨牌有多少种不同的放法。这是一种二维前缀和。放骨牌的话只有两种放法横着和竖着。那么可以设两个二维数组分别装横与竖着的前缀数组。然后再枚举两个符合的区间。

#include<cstdio>
#include <cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#define MX 501
using namespace std;

int h,w,D[MX][MX],W,R[MX][MX],DD[MX][MX],RR[MX][MX];
char A[MX][MX];

int main()
{
int i,m,a,b,c,d,j;
scanf("%d%d",&h,&w);
for(i=0;i<h;i++)
{
scanf("%s",A[i]);
}
for(i=0;i<h;i++)
{
for(j=0;j<w;j++)
{
W=0;
if(i>=1&&A[i][j]=='.'&&A[i-1][j]=='.')
{
W++;
}
D[i+1][j+1]=W;
W=0;
if(j>=1&&A[i][j]=='.'&&A[i][j-1]=='.')
{
W++;
}
R[i+1][j+1]=W;
}
}
for(i=1;i<=h;i++)
{
for(j=1;j<=w;j++)
{
DD[i][j]=D[i][j]+DD[i][j-1]+DD[i-1][j]-DD[i-1][j-1];
RR[i][j]=R[i][j]+RR[i][j-1]+RR[i-1][j]-RR[i-1][j-1];
}
}
scanf("%d",&m);
for(i=0;i<m;i++)
{
scanf("%d%d%d%d",&a,&b,&c,&d);
printf("%d\n",RR[c][d]-RR[c][b]-RR[a-1][d]+RR[a-1][b]+DD[c][d]-DD[c][b-1]-DD[a][d]+DD[a][b-1]);
}
return 0;
}