Codeforces Round #343 (Div. 2) A. Far Relative’s Birthday Cake【暴力/组合数】

A. Far Relative’s Birthday Cake

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!

The cake is a n × n square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?

Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.

Input

In the first line of the input, you are given a single integer n (1 ≤ n ≤ 100) — the length of the side of the cake.

Then follow n lines, each containing n characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.

Output

Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.

Examples

input

3
.CC
C..
C.C

output

4

input

4
CC..
C..C
.CC.
.CC.

output

9

Note

If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:

1. (1, 2) and (1, 3)
2. (3, 1) and (3, 3)

Pieces that share the same column are:

1. (2, 1) and (3, 1)
2. (1, 3) and (3, 3)

【题意】：给出一个n*n的蛋糕，’C’表示巧克力，’.’表示空，如果任一行或者任一列有两个巧克力，小明的快乐度会加一，问小明的快乐度是多少 。

【分析】：实际就是，问同一列两个一组的组合数和同一行两个一组和组合数的和。

【代码】：

#include <iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<streambuf>
#include<cmath>
#include<string>
using namespace std;
#define ll long long
#define oo 10000000
char a[500][500];
int cnt;
int sum;
int n;
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
       scanf("%s",&a[i]);
       sum=0;
    for(int i=0;i<n;i++)
    {
        cnt=0;
        for(int j=0;j<n;j++)
        {
            if(a[i][j]=='C')
            {
                cnt++;
            }
        }
        sum+=(cnt)*(cnt-1)/2;
    }

    for(int j=0;j<n;j++)
    {
        cnt=0;
        for(int i=0;i<n;i++)
        {
            if(a[i][j]=='C')
            {
                cnt++;
            }
        }
        sum+=(cnt)*(cnt-1)/2;
    }
    printf("%d
",sum);
    return 0;
}