Codeforces AIM Tech Round 5 (rated, Div. 1 + Div. 2)

A. Find Square

time limit per test:

1 second

memory limit per test:

256 megabytes

input:

standard input

output:

standard output

Consider a table of size n×mn×m, initially fully white. Rows are numbered 11 through nn from top to bottom, columns 11 through mm from left to right. Some square inside the table with odd side length was painted black. Find the center of this square.

Input

The first line contains two integers nn and mm (1≤n,m≤1151≤n,m≤115) — the number of rows and the number of columns in the table.

The ii-th of the next nn lines contains a string of mm characters si1si2…simsi1si2…sim (sijsij is 'W' for white cells and 'B' for black cells), describing the ii-th row of the table.

Output

Output two integers rr and cc (1≤r≤n1≤r≤n, 1≤c≤m1≤c≤m) separated by a space — the row and column numbers of the center of the black square.

Examples

input

5 6
WWBBBW
WWBBBW
WWBBBW
WWWWWW
WWWWWW

output

2 4

input

3 3
WWW
BWW
WWW

output

2 1

Solution1:

#include <iostream>
const int N = 712;
char maze[N][N];
int a, b, r, c, n;
using namespace std;
void dfs(int i, int k)
{
	if (maze[i][k] == 'W') maze[i][k] = 'w';
	else
	{
		r += i, c += k; n += 1;
		maze[i][k] = 'b';
	}
	for (int j = 0; j < b; j++)
		if (maze[i][j] == 'W' || maze[i][j] == 'B')
			dfs(i, j);
}
int main()
{
	scanf_s("%d %d", &a, &b);
	r = c = n = 0;
	for (int i = 0; i < a; i++)
		for (int j = 0; j < b; j++)
			cin >> maze[i][j];
	for (int i = 0; i < a; i++)
		if (maze[i][0] == 'W' || maze[i][0] == 'B')
			dfs(i, 0);

	if (n != 0)
	{
		r /= n, c /= n;
		printf("%d %d\n", r + 1, c + 1);
	}
	return 0;
}

Solution2:

#include <iostream>
const int N = 712;
char maze[N][N];
int a, b, r, c, n;
using namespace std;
int main()
{
	r = c = n = 0;
	scanf_s("%d %d", &a, &b);
	for (int i = 0; i < a; i++)
		for (int j = 0; j < b; j++)
		{
			cin >> maze[i][j];
			if (maze[i][j] == 'B')
				r += i, c += j, n += 1;
		}
	if (n != 0)
	{
		r /= n, c /= n;
		printf("%d %d\n", r + 1, c + 1);
	}
	return 0;
}