Look at the following partial schedule involving four transactions T$1,$ T$2,$ T$3$ and T$4$; and the data items A$,$ B$,$ and C$.$

Partial Schedule:

T$4\_$lock$\_$X$($A$)$; T$4\_$R$($A$)$; T$1\_$lock$\_$X$($B$)$; T$1\_$R$($B$)$; T$4\_$W$($A$)$; T$3\_$lock$\_$S$($C$)$; T$3\_$R$($C$)$; T$2\_$lock$\_$S$($B$)$; T$2\_$R$($B$)$; T$2\_$lock$\_$S$($C$)$; T$4\_$lock$\_$S$($B$)$; T$3\_$releaseLock$($C$)$; T$1\_$W$($B$)$;T$1\_$releaseLock$($B$)$; T$4\_$releaseLock$($A$)$;

Note:

T$2\_$lock$\_$S$($A$)$ means T$2$ locks data item A in Share mode.

T$1\_$lock$\_$X$($B$)$ means T$1$ locks data item B in Exclusive mode.

T$2\_$R$($A$)-$ means T$2$ reads data item A

T$3\_$W$($A$)-$ means T$3$ writes data item A

T$2\_$releaseLock$($B$)-$ means T$2$ release lock on data item B

Check if the above partial schedule results in a deadlock by drawing a wait$-$for graph. Give your reasoning. Give the final wait$-$for graph.