Count occurrences of each digit in its row and column

Given a matrix $u$ of digits, we have to produce a new matrix $v$ of digits with the same dimensions, where the digit in each position is calculated in the following way.

Let $i,j$ be a position in $u$ , and let $d$ be the digit in $u[i][j]$ . Then, $v[i][j]$ is the number of occurrences of $d$ in $u$ within row $i$ , plus the number of occurrences of $d$ in $u$ within column $j$ , and all of it modulus $10$ so that the result is again a digit.

Note that the occurrence of $d$ at position $i,j$ will be counted twice, one for row $i$ and another for column $j$ .

For instance, consider this matrix $u$ :

Then, the resulting matrix $v$ is:

For instance, $u[1][1] = 2$ . At row $1$ we have only one $2$ . And at column $1$ we have two $2$ . Therefore $v[1][1] = (1+2)\%10 = 3$ .

Input

The input has several cases. Each case starts with two positive natural numbers, $n, m$ , in one line. After that comes a matrix of $n\times{}m$ digits ( $n$ lines with $m$ digits each), followed by an empty line.

Output

For each case, the program has to write $n$ lines with $m$ digits each, which is the resulting matrix, followed by an empty line.

Observation

Evaluation out of 10 points:

Slow solution: 5 points.
Fast solution: 10 points.

A fast solution is one which is correct, of linear cost and passing the test cases, both public and private. A slow solution is one which is not fast, but it is correct and passes the public test cases.

Problem information

Author: PRO1

Generation: 2026-01-25T22:38:39.288Z