Problem:
Score : 100 points

Problem Statement

We have a 3×3 square grid, where each square contains a lowercase English letters.
The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}.

Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.

Constraints

Input consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

c_{11}c_{12}c_{13}
c_{21}c_{22}c_{23}
c_{31}c_{32}c_{33}

Output

Print the string of length 3 that can be obtained by concatenating the letters on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.

Sample Input 1

ant
obe
rec

Sample Output 1

abc

The letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid are a, b and c from top-right to bottom-left. Concatenate these letters and print abc.

Sample Input 2

edu
cat
ion

Sample Output 2

ean

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main()
{
	int map[1000][100];
	int m, n;
	int i, j;
	scanf("%d",&m);
	scanf("%d",&n);
	for (i = 0; i < m; i++)
	{
		for (j = 0; j < n; j++)
		{
			map[i][j] = 1;
		}
	}

	for (i = 0; i < m; i++)
	{
		for (j = 0; j < n; j++)
		{
			map[i][j] = -map[i][j];
			map[i][j + 1] = -map[i][j + 1];
			map[i + 1][j] = -map[i + 1][j];
			map[i + 1][j + 1] = -map[i + 1][j + 1];

			if (i - 1 != -1)
			{
				map[i - 1][j] = -map[i - 1][j];
				map[i - 1][j + 1] = -map[i - 1][j + 1];
			}
			if (j - 1 != -1)
			{
				map[i][j - 1] = -map[i][j - 1];
				map[i + 1][j - 1] = -map[i + 1][j - 1];
			}
			if (i - 1 != -1 && j - 1 != -1)
			{
				map[i - 1][j - 1] = -map[i - 1][j - 1];
			}
		}
	}

	int count = 0;
	for (i = 0; i < m; i++)
	{
		for (j = 0; j < n; j++)
		{
			if (map[i][j] == -1)
				count++;
		}
	}
	printf("%d",count);
	return 0;
}