B. Little Elephant and Magic Square

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Little Elephant loves magic squares very much.

A magic square is a 3 × 3 table, each cell contains some positive integer. At that
the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15.



The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements
of the magic square did not exceed 105.

Help the Little Elephant, restore the original magic square, given the Elephant's notes.

Input

The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements
that have been forgotten by the Elephant are represented by zeroes.

It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.

Output

Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105.

It is guaranteed that there exists at least one magic square that meets the conditions.

Sample test(s)

input

0 1 1
1 0 1
1 1 0

output

1 1 1
1 1 1
1 1 1

input

0 3 6
5 0 5
4 7 0

output

6 3 6
5 5 5
4 7 4

解题说明：此题要求行，列，对角线数字之和都相等，而且丢失的元素位置固定了，等于是解一个方程组了，完全不需要暴力

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;

int main()
{
int A,B,C,D,E,F,G,H,I;
scanf("%d%d%d%d%d%d%d%d%d",&A,&B,&C,&D,&E,&F,&G,&H,&I);
I = D + (F - H)/2;
A = (F+H)/2;
E = B+C - I;
printf("%d %d %d\n%d %d %d\n%d %d %d\n",A,B,C,D,E,F,G,H,I);
return 0;
}