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#banana #cantor #joe #man #rob #tie
Published: 2015-01-01 20:11:58 +0000 UTC; Views: 464; Favourites: 23; Downloads: 0
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Description
HAPPY NEW YEARRRRR !!this here is Nana and he is a banana
(he might secretly be rob cantor. who knows?) I for some reason love drawing bananas ???
happy 2015 may all of your dreams come trueee
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Comments: 27
mothkit [2015-02-14 00:59:16 +0000 UTC]
i remember this was the first art i ever saw from you
amazing
π: 0 β©: 1
sailrs [2015-01-23 23:58:24 +0000 UTC]
if joe H. is in tally hall then i have to admit
joehawley robcantor tallyhall
π: 0 β©: 1
chmpr In reply to sailrs [2015-01-24 00:13:48 +0000 UTC]
user robcantor is actually my friend. B""00 joe hawley is indeed in tally hall, but I don't think dA user tallyhall is real.
π: 0 β©: 1
1DRusherGamer [2015-01-19 04:31:01 +0000 UTC]
Banana man... HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA
π: 0 β©: 1
chmpr In reply to Drill8Bit [2015-01-07 02:57:21 +0000 UTC]
I FEEL LIKE A SHY ENORMOUS PINK FLAMINGO MANNNNNNNNnn
FLAMINGOOOOOOOOOOOOOOOOOOo
but really this is for banana man.
BOTH SONGS ARE BEYOND AMAZING THOUGH. I might draw rob as a flamingo man, actually.
π: 0 β©: 1
Drill8Bit In reply to chmpr [2015-01-07 11:42:45 +0000 UTC]
Dammit - I'm sucked into rob's vortex.
π: 0 β©: 1
chmpr In reply to Drill8Bit [2015-01-07 18:00:47 +0000 UTC]
no one shall escape rob's vortex
his music will put a lasso around your neck and pull you right in
π: 0 β©: 0
chmpr In reply to EmilyVanSlyke [2015-01-01 22:45:38 +0000 UTC]
you could say you're 'bananas' for him
laughs
π: 0 β©: 1
EmilyVanSlyke In reply to chmpr [2015-01-01 22:46:45 +0000 UTC]
pfft
lets peel his banana
π: 0 β©: 1
chmpr In reply to EmilyVanSlyke [2015-01-01 22:56:00 +0000 UTC]
ohno his true identity
he's veryΒ self conscious about it
π: 0 β©: 1
chmpr In reply to let-us-sing [2015-01-01 20:27:48 +0000 UTC]
this banana 4 u
B"")))))))))))))
π: 0 β©: 1
let-us-sing In reply to chmpr [2015-01-01 20:35:15 +0000 UTC]
i showed that music video to my friend once and now she's obsessed with it XD
π: 0 β©: 1