C. Target Practice

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A $10×10$ target is made out of five "rings" as shown. Each ring has a different point value: the outermost ring — 1 point, the next ring — 2 points, ..., the center ring — 5 points.

Vlad fired several arrows at the target. Help him determine how many points he got.

Input

The input consists of multiple test cases. The first line of the input contains a single integer $t$ ($1≤t≤1000$) — the number of test cases.

Each test case consists of 10 lines, each containing 10 characters. Each character in the grid is either $X$ (representing an arrow) or $.$ (representing no arrow).

Output

For each test case, output a single integer — the total number of points of the arrows.

Example

input

Copy

4

X.........

..........

.......X..

.....X....

......X...

..........

.........X

..X.......

..........

.........X

..........

..........

..........

..........

..........

..........

..........

..........

..........

..........

..........

..........

..........

..........

....X.....

..........

..........

..........

..........

..........

XXXXXXXXXX

XXXXXXXXXX

XXXXXXXXXX

XXXXXXXXXX

XXXXXXXXXX

XXXXXXXXXX

XXXXXXXXXX

XXXXXXXXXX

XXXXXXXXXX

XXXXXXXXXX

output

Copy

17
0
5
220

Note

In the first test case, there are three arrows on the outer ring worth 1 point each, two arrows on the ring worth 3 points each, and two arrows on the ring worth 4 points each. The total score is $3×1+2×3+2×4=17$.

In the second test case, there aren't any arrows, so the score is $0$.

3

解题说明：此题是一道几何题，图从外到里依次为1-5分，X为射中了这个点，求总分。设某点坐标x,y，则该点分值为 { x，y，10-x+1，10-y+1 } 取其中的最小值，对所有射中点求和即可。

#include
int main()
{
	int t;
	scanf("%d", &t);
	char c;
	scanf("%c", &c);
	for (int i = 0; i < t; i++)
	{
		int s = 0;
		for (int j = 0; j < 10; j++) 
		{
			for (int k = 0; k < 10; k++) 
			{
				scanf("%c", &c);
				if (c == 'X')
				{
					int a = 5;
					if (j + 1 < a)
					{
						a = j + 1;
					}
					if (10 - j < a)
					{
						a = 10 - j;
					}
					if (k + 1 < a)
					{
						a = k + 1;
					}
					if (10 - k < a)
					{
						a = 10 - k;
					}
					s += a;
				}
			}
			scanf("%c", &c);
		}
		printf("%d\n", s);
	}
	return 0;
}