From Wikipedia:

In modular arithmetic,   the modular multiplicative inverse of an integer    a    modulo    m    is an integer    x    such that





a

x
≡
1


(
mod

m
)

.


{\displaystyle a\,x\equiv 1{\pmod {m}}.}


Or in other words, such that:





∃
k
∈

Z

,

a

x
=
1
+
k

m


{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}


It can be shown that such an inverse exists   if and only if    a    and    m    are coprime,   but we will ignore this for this task.



Task

Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language,   compute the modular inverse of   42 modulo 2017.




MODULE ModularInverse;
  FROM InOut IMPORT WriteString, WriteInt, WriteLn;
 
  TYPE Data = RECORD x : INTEGER;
                     y : INTEGER
              END;
 
  VAR c  : INTEGER;
      ab : ARRAY [1..5] OF Data;
 
PROCEDURE mi(VAR a, b : INTEGER): INTEGER;
  VAR t, nt, r, nr, q, tmp : INTEGER;
 
BEGIN
  b := ABS(b);
  IF a < 0 THEN a := b - (-a MOD b) END;
  t := 0; nt := 1; r := b; nr := a MOD b;
  WHILE (nr # 0) DO
    q := r / nr;
    tmp := nt; nt := t - q * nt; t := tmp;
    tmp := nr; nr := r - q * nr; r := tmp;
  END;
  IF (r > 1) THEN RETURN -1 END;
  IF (t < 0) THEN RETURN t + b END;
  RETURN t;
END mi;
 
BEGIN
  ab[1].x := 42;   ab[1].y := 2017;
  ab[2].x := 40;   ab[2].y := 1;
  ab[3].x := 52;   ab[3].y := -217;
  ab[4].x := -486; ab[4].y := 217;
  ab[5].x := 40;   ab[5].y := 2018;
  WriteLn;
  WriteString("Modular inverse");
  WriteLn;
  FOR c := 1 TO 5 DO
    WriteInt(ab[c].x, 6); WriteString(", ");
    WriteInt(ab[c].y, 6); WriteString(" = ");
    WriteInt(mi(ab[c].x, ab[c].y),6);
    WriteLn;
  END;
END ModularInverse.