Given below is a directed acyclic graph where each arc is separated by a semicolon ";". Each arc is represented by four tokens which are separated by a space character. First token is written as start-node -> end-node and it represents that the arc connects startnode to the endnode. Second token represents the label on the arc.

The graph is:
0->1 silence ; 1->2 call ; 2->3 NAME_1 ; 2->3 NAME_2 ; 2->3 NAME_3 ; 2->3 Ambar ; 2->3 NAME_4 ; 2->3 Amber ; 3->4 Sadek ; 3->4 Sadiq; 3->4 Sodiq ; 3->4 NAME_5; 3->4 Sodek; 3->4 Sadik ; 3->4 Saddiq ; 3->4 Sodak ; 4->5 silence; end-node=5

If multiple arcs have same start-node and end-node, it means that they are connecting the same node. Compress all such arcs to one single arc with the same start-node and end-node with all labels concatenated together by a backslash character. For example, the arc "1->2 abc" and arc "1->2 def" can be compressed as "1->2 abc/def" because they have the same start-node "1" and end-node "2" respectively.

Compress the above graph for me. Classify arc as "confidence" if it has only one label. Classify arc as "confused" if it has multiple labels separated by backslash character "/".

If most the arcs are classified as "confident", label the graph as "1"
If most the arcs are classified as "confused", label the graph as "0" Make sure to only use capital letters in your entire response.
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