Time series (yearly figures) below and I use R to fit it with ARIMA

library(forecast)

beer <- c(150,241,361,403,504,684,706,862,879,806,840,846,1024,1196,1239,1237,1281,1342)

ts_beer = ts(beer, start = c(1980), frequency = 1)

dif.ts_beer <- diff(ts_beer)

acf(dif.ts_beer)
pacf(dif.ts_beer)

Based on the ACF and PACF, I make it ARIMA(4,0,4)

dif.ts_beer.fit <- arima(dif.Gas, order = c(4,0,4))

dif.ts_beer.fit

It looks ok. But when I run auto.arima:

auto.arima(dif.ts_beer)

It gives:

Series: dif.ts_beer 
ARIMA(0,0,0) with non-zero mean 

Coefficients:
         mean
      70.1176
s.e.  17.0359

sigma^2 estimated as 5242:  log likelihood=-96.4
AIC=196.81   AICc=197.67   BIC=198.48

So the manual ARIMA(4,0,4) is not a good choice for this case? If so, what parameters shall be given to the ARIMA()?

Thank you.