Free refined value in Rietveld I have a question about Rietveld refinement

The author of this paper claimed that the free refined CN ligand is 1.081. 
However, they does not show how they got that number. They stated it was based from this table however I am not sure how you can get 1.081 from this table. What is the calculation?



Thank you.

Source: 

Reference


Wang, W.; Gang, Y.; Hu, Z.; Yan, Z.; Li, W.; Li, Y.; Gu, Q.-F.; Wang, Z.; Chou, S.-L.; Liu, H.-K.; Dou, S.-X. Reversible Structural Evolution of Sodium-Rich Rhombohedral Prussian Blue for Sodium-Ion Batteries. Nature Communications 2020, 11 (1), 980. DOI: 10.1038/s41467-020-14444-4
I guess that 1.081 refers to the distance length (in Angstrom) separating the C and N atoms. You obtain this value by applying by using the values listed in the table; this is quite tedious, takes some time and require quite good knowledge about crystallography. Otherwise (and much more easy), you can use the crystallographic data listed in the table to build the structure using a proper crystallographic software (such as Vesta); then the software will calculate bond distances for you.