(x,y) = (m_1,n_1) is the least positive solution of bx-ay= 1 while performing euclidean algorithm

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I was reading number theory book by John Stillwell and I am stuck somewhere.

The symbolic Euclidean algorithm is used when solving linear Diophantine equations. Suppose that we run the ordinary Euclidean algorithm on the numbers $b$ and $-a$ until $1$ and $-1$ are produced, and suppose that the corresponding vector pair is $((m_1 , n_1 ), (m_2 ,n_2))$.

It claims that $(x,y) = (m_1,n_1)$ is the least positive solution of $bx-ay= 1$ and that $(x,y)= (m_2,n_2)$ is the least positive solution of $bx-ay=-1$.

It certainly looks like that but I got no clue to solve it. enter image description here enter image description here