T (B)
OLL parity (1 flip)
2 Corners
Count the flipped dedges (one here), then read the corner shape as for OLL.
Your status
Algorithms (3)
(U2) first, then:
3Rw' D R2 Rw2 B2 Rw' U2 Rw' U2 x' U2 Rw' U2 Rw U2 Rw'
U2 Rw2 U2 x U2 R2 D' 3Rw
My choice
22 HTM · 22 STM · 35 QTM
rotation
<U,R,D,B>
One hand: needs L, B, S or E
Do U2 first (AUF) on the case as drawn. · Cubemuch
Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw'
U2 Rw' U R U2 R2 U' R2 U' R2 U2 R
27 HTM · 27 STM · 40 QTM
3-gen
rotation
<U,R,L>
One hand: needs L, B, S or E
Cubemuch
L B' R2 R 2R R 2R B2 2R' R' U2 2R' R' U2 B2 2R'
R' B2 R 2R B2 2R' R' B2 R 2R R 2R B2 U2 R2 B
L'
42 HTM · 33 STM · 53 QTM
rotationless
<U,R,L,B>
One hand: needs L, B, S or E
Cubemuch
How this case is verified
The case is the source's setup (or, where it has none, another source algorithm reversed). Setup: R' U2 R2 U R2 U R2 U2 R' U' Rw U2 Rw U2 Rw' U2 Rw U2 Lw' U2 Rw U2 Rw' U2 Rw' x' U2 Rw'. Every algorithm above was run by our logic engine from that state and reaches "oll-solved", keeping first-layers.