New Year and Domino 二维前缀和-优快云博客

本文介绍了一个关于放置多米诺骨牌的算法问题，包括输入输出格式、问题描述及解决方案。通过预处理空闲网格，快速计算指定矩形区域内放置多米诺骨牌的不同方式数量。

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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 1through 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

Copy

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

output

Copy

input

Copy

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

Copy

Note

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

　　自闭自闭

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <set>
 7 #include <iostream>
 8 #include <map>
 9 #include <stack>
10 #include <string>
11 #include <vector>
12 #define  pi acos(-1.0)
13 #define  eps 1e-6
14 #define  fi first
15 #define  se second
16 #define  lson l,m,rt<<1
17 #define  rson m+1,r,rt<<1|1
18 #define  bug         printf("******\n")
19 #define  mem(a,b)    memset(a,b,sizeof(a))
20 #define  fuck(x)     cout<<"["<<x<<"]"<<endl
21 #define  f(a)        a*a
22 #define  sf(n)       scanf("%d", &n)
23 #define  sff(a,b)    scanf("%d %d", &a, &b)
24 #define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
25 #define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
26 #define  pf          printf
27 #define  FRE(i,a,b)  for(i = a; i <= b; i++)
28 #define  FREE(i,a,b) for(i = a; i >= b; i--)
29 #define  FRL(i,a,b)  for(i = a; i < b; i++)
30 #define  FRLL(i,a,b) for(i = a; i > b; i--)
31 #define  FIN         freopen("DATA.txt","r",stdin)
32 #define  gcd(a,b)    __gcd(a,b)
33 #define  lowbit(x)   x&-x
34 #pragma  comment (linker,"/STACK:102400000,102400000")
35 using namespace std;
36 typedef long long LL;
37 typedef unsigned long long ULL;
38 const int maxn = 2e5 + 10;
39 int n, m, sum[505][505], y[505][505], x[505][505];
40 char mp[505][505];
41 int main() {
42     sff(n, m);
43     for (int i = 1 ; i <= n ; i++)
44         scanf("%s", mp[i] + 1);
45     for (int i = 1 ; i <= n ; i++) {
46         for (int j = 1 ; j <= m ; j++) {
47             if (mp[i][j] == '.' && mp[i + 1][j] == '.' ) {
48                 sum[i][j]++;
49                 y[i][j] = 1;
50             }
51             if (mp[i][j] == '.' && mp[i][j + 1] == '.' ) {
52                 sum[i][j]++;
53                 x[i][j] = 1;
54             }
55             sum[i][j] += sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
56         }
57     }
58     int q;
59     sf(q);
60     while(q--) {
61         int x1, y1, x2, y2;
62         sffff(x1, y1, x2, y2);
63         int ans = sum[x2][y2] - sum[x1 - 1][y2] - sum[x2][y1 - 1] + sum[x1 - 1][y1 - 1];
64         for (int i = x1 ; i <= x2 ; i++) ans -= x[i][y2];
65         for (int i = y1 ; i <= y2 ; i++) ans -= y[x2][i];
66         printf("%d\n", ans);
67     }
68     return  0;
69 }