codeforces 828B. Black Square（水题）

题目链接:http://codeforces.com/contest/828/problem/B点击打开链接

B. Black Square

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp has a checkered sheet of paper of size n × m. Polycarp painted some of cells with black, the others remained white. Inspired by Malevich's "Black Square", Polycarp wants to paint minimum possible number of white cells with black so that all black cells form a square.

You are to determine the minimum possible number of cells needed to be painted black so that the black cells form a black square with sides parallel to the painting's sides. All the cells that do not belong to the square should be white. The square's side should have positive length.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the sizes of the sheet.

The next n lines contain m letters 'B' or 'W' each — the description of initial cells' colors. If a letter is 'B', then the corresponding cell is painted black, otherwise it is painted white.

Output

Print the minimum number of cells needed to be painted black so that the black cells form a black square with sides parallel to the painting's sides. All the cells that do not belong to the square should be white. If it is impossible, print -1.

Examples

input

5 4WWWWWWWBWWWBWWBBWWWW

output

5

input

1 2BB

output

-1

input

3 3WWWWWWWWW

output

1

Note

In the first example it is needed to paint 5 cells — (2, 2), (2, 3), (3, 2), (3, 3) and (4, 2). Then there will be a square with side equal to three, and the upper left corner in (2, 2).

In the second example all the cells are painted black and form a rectangle, so it's impossible to get a square.

In the third example all cells are colored white, so it's sufficient to color any cell black.

题意：让你画全黑的正方形 要求其余地方是白的 求出这个正方形的四个角边界 判断是否超出原有图形 如果都没有边界则输出1

#include <stdio.h>#include <stdlib.h>#include <iostream>#include<algorithm>#include <math.h>#include <string.h>#include <limits.h>#include <string>#include <queue>#include <stack>#include <set>#include <vector>using namespace std;char a[111][111];int minx=INT_MAX,miny=INT_MAX,maxx=0,maxy=0;int main(){    int n,m;int sum=0;    scanf("%d%d",&n,&m);    getchar();    for(int i=1;i<=n;i++)    {        for(int j=1;j<=m;j++)        {            scanf("%c",&a[i][j]);            if(a[i][j]=='B')            {                minx=min(minx,i);                maxx=max(maxx,i);                miny=min(miny,j);                maxy=max(maxy,j);                sum++;            }        }        getchar();    }    //printf("%d %d %d %d\n",minx,maxx,miny,maxy);    int maxedge=0;    if(!sum)    {        printf("1");        return 0;    }    maxedge=max(maxx-minx+1,maxy-miny+1);    if(maxedge>n||maxedge>m)        printf("-1");    else        printf("%d",maxedge*maxedge-sum);}