Takahashi is initially at (0, 0)(0,0) facing east. Then, he moves as follows.
-
t_1 =t1= S, so he advances in the direction of east by distance 11, arriving at (1, 0)(1,0).
-
t_2 =t2= S, so he advances in the direction of east by distance 11, arriving at (2, 0)(2,0).
-
t_3 =t3= R, so he turns 9090 degrees clockwise, resulting in facing south.
-
t_4 =t4= S, so he advances in the direction of south by distance 11, arriving at (2, -1)(2,−1).
Thus, Takahashi's final position, (x, y) = (2, -1)(x,y)=(2,−1), should be printed.