function fib(n) {
  // check for simple case 
  if (n === 0) return 0n;
  if (n === 1 || n === -1) return 1n;
  
  // handle negative signs for negative n
  let sign = 1n; 
  if (n < 0) {
    // change n to positive for the recursive nthPowerMatrixF
    n *= -1;
    // if negative n is even, we will return F-n = -Fn
    if (-n % 2 === 0) {
      sign = -1n;
    }
  }
  
  /* using Fibonacci matrix approach
  the matrix F below shows the reccurence relation of the
  fibonacci numbers */
  const F = [
    [1n, 1n],
    [1n, 0n]
  ]
  
  
  // helper function to multiply two 2 x 2 matrices
  function multiplyMatrices(F, A) {
    // @params F is the Fibonocci matrix
    // @params A the second matrix to multiply with F.
    // @return return nothing as this function will modify F
    const a = (F[0][0] * A[0][0]) + (F[0][1] * A[1][0]);
	  const b = (F[0][0] * A[0][1]) + (F[0][1] * A[1][1]);
	  const c = (F[1][0] * A[0][0]) + (F[1][1] * A[1][0]);
	  const d = (F[1][0] * A[0][1]) + (F[1][1] * A[1][1]);
    
    F[0][0] = a;
    F[0][1] = b;
    F[1][0] = c;
    F[1][1] = d;
  }
  
  // define nthPowerMatrixF which will be a recursive function
  function nthPowerMatrixF(F, n) {
    // base case of recursive
    if (n === 0 || n === 1) return;
    
    // as we will mutate F, we need a local matrix to represent the
    // original F matrix
    const A = [
      [1n, 1n],
      [1n, 0n]
    ]
    // if n is odd
    if (n % 2 !== 0) {
      nthPowerMatrixF(F, n - 1);
      multiplyMatrices(F,A);
      return;
    }
    
    // if n is even, we can work half n to do less calculations
    // e.g n = 4 we only need to do n = 2, then multiply by it self to get n = 4
    nthPowerMatrixF(F, n / 2);
    multiplyMatrices(F, F);
    
  }
  
  // formula states [Fn, Fn-1] = (F to the power of n - 1) * [1,0]
  nthPowerMatrixF(F, n - 1);
  
  // resultant matrix is [[Fn, Fn-1,], [Fn-1, Fn-2]]
  return sign * F[0][0];
}