Good Bye C. New year and Domino two-dimensional prefix

C. New Year and Domino

They say "years is 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 was a square, either empty (denoted by ') or forbidden (denoted by '# '). Rows is numbered 1 through H from top to bottom. Columns is numbered1 through w from left to right.

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

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

Input

The first line of the input contains, integers h and w (1≤ H, w ≤500) , haven 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 are either '. ' or '# '-denoting an empty or forbidden cell, respectively.

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

Each of the nextQLines contains four integersR1I,C1I,R2I,C2 i   (1≤ R 1 I ≤ R 2 i ≤ h , 1≤ c 1 I ≤ c 2 i ≤ w )  -the  I -th Query. Numbers  R 1 i  and C 1 i  denote the row and the column ( respectively) of the upper left cell of the rectangle. Numbers  R 2 i  and  C 2 i  denote the row and the column ( respectively) of the bottom right cell of the rectangle.

Output

Print q integers, i-th should is equal to the number of ways to put a single domino inside the c5>i-th rectangle.

Sample Test (s) 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 is placed in 4 possible ways.

Test instructions: Give you a picture of n*m, there. , #, #号不可走

Q a Query

Each query for a rectangle asks how many paths in this rectangle are the length 2

Puzzle: We can find out if each point can go down to the right value is 2,1,0,

Then from 1, 1 to x, y this rectangle can be obtained with a two-dimensional prefix array

The answer to each question is simple, with a focus on the boundary and only two directions.

//Meek#include <bits/stdc++.h>#include<iostream>#include<cstdio>#include<cmath>#include<string>#include<cstring>#include<algorithm>#include<map>#include<queue>using namespacestd; typedefLong Longll;#defineMem (a) memset (A,0,sizeof (a))#definePB Push_back#defineFi first#defineSe Second#defineMP Make_pairConst intn=550+ -;Constll INF = 1ll<< A;Const intINF =1000000007;Const intMod=1000007;CharMp[n][n];intP[n][n],v[n][n];intn,m;intsss[4][2] ={1,0,0,1};structss{intx, y;};intCheckintXinty) {if(x<=0|| y<=0|| x>n| | Y&GT;M)return 1; return 0;}intV[n][n];voidBFsintXinty) { for(intI=1; i<=n;i++) {         for(intj=1; j<=m;j++) {                if(mp[i][j]!='#')              for(intk=0;k<2; k++) {                 intxx = i+sss[k][0]; intyy = j+sss[k][1]; if(check (XX,YY))Continue; if(mp[xx][yy]=='#')Continue; P[I][J]++; }        }    }}intMain () {scanf ("%d%d",&n,&m);  for(intI=1; i<=n;i++) {GetChar ();  for(intj=1; j<=m;j++) {scanf ("%c",&Mp[i][j]); }} BFS (1,1);  for(intI=1; i<=n;i++) {         for(intj=1; j<=m;j++) {V[i][j]= v[i][j]+v[i-1][j]+v[i][j-1]-v[i-1][j-1]+P[i][j]; }     }    intQ,x,y,x2,y2; scanf ("%d",&q);  for(intI=1; i<=q;i++) {scanf ("%d%d%d%d",&x,&y,&x2,&y2); intAns = v[x2][y2]-v[x2][y-1]-v[x-1][y2]+v[x-1][y-1];  for(intj=x;j<x2;j++) {            if(mp[j][y2]=='.'&&mp[j][y2+1]=='.') ans--; } ans-=P[x2][y2];  for(intj=y;j<y2;j++) {            if(mp[x2][j]=='.'&&mp[x2+1][j]=='.') ans--; } printf ("%d\n", ans); }}

Code

Good Bye C. New year and Domino two-dimensional prefix