Pi (F)
OLL parity (1 flip)
0 Corners
Count the flipped dedges (one here), then read the corner shape as for OLL.
Your status
Algorithms (2)
(U) first, then:
z' U2 Lw' R2 x U' F2 U Lw' U2 2R' U2 Rw F' L2 F
2R2 U2 x' U' F2 U x U2 Lw' U2 z
My choice
24 HTM · 22 STM · 36 QTM
rotation
<U,R,F,L>
One hand: needs L, B, S or E
Do U 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
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.