FFTW_BACKWARD: Every other element looks bad

I use the following code to get an RFT of an 8 byte complex data
array.  sPadding=1 in this case.

typedef struct Complex_Struct
{
	float	fReal;
	float	fImaginary;
} Complex;

ERROR_NUMBER CMyLib::FFTWMake1DInverseFourierTransform(Complex
*cpInputVector, float **fppOutputVector,    DWORD dwVectorLength,
short sPadding)
{
    fftw_complex *in, *out;
    fftw_plan p;
	size_t	stNumElements=dwVectorLength*sPadding;
	size_t		iInc;

	if (dwVectorLength<1 || sPadding<1) return ERROR_PARAMETERS;

    if (!(in = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) *
stNumElements)) ||
		!(out = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) *
stNumElements)))
	{
		// Error handling
	}

	// Create plan.
	if (!(p = fftw_plan_dft_1d((int)stNumElements, in, out,
FFTW_BACKWARD, FFTW_EXHAUSTIVE)))
	{
		// Error handling
	}

	// Initialise input elements to zero if required
	if (sPadding>1)
	{
		for (iInc=dwVectorLength; iInc<stNumElements; ++iInc) in[iInc]
[0]=in[iInc][1]=0;
	}
	for (iInc=0; iInc<dwVectorLength; ++iInc)
	{
		in[iInc][0]=cpInputVector[iInc].fReal;
		in[iInc++][1]=cpInputVector[iInc].fImaginary;
	}

	// Do FFT.
	fftw_execute(p); // repeat as needed

	// Free input fftw_complex vector
	fftw_destroy_plan(p);
	fftw_free(in);

	// Allocate memory to function output
	if (!(*fppOutputVector=(float *)malloc(stNumElements*sizeof(float))))
	{
		// Error handling
	}

	// Copy fftw_complex out vector into function output vector
	for (iInc=0; iInc<stNumElements; ++iInc)
	{
		(*fppOutputVector)[iInc]=(float)(out[iInc][0]);
	}

                // Etc.
}

I get the following output.

	out[0][0]	1636200805.3022504	double
	out[1][0]	-3.7190885739783542e-007	double
	out[2][0]	1635147705.0245562	double
	out[3][0]	-2.0041953678394723e-007	double
	out[4][0]	1634009932.1876054	double
	out[5][0]	-3.1898067317028498e-007	double


It would appear that the even numbered elements are what I might
expect but the odd numbered elements are wrong.

Any assistance would be greatly appreciated.

Many thanks in advance,
Peter.