A friend of mine runs a pharmacy.   He has a specialized function in his Dispensary application which receives a decimal value of currency and replaces it to a standard value.   This value is regulated by a government department.



Task

Given a floating point value between   0.00   and   1.00,   rescale according to the following table:

>=  0.00  <  0.06  :=  0.10
>=  0.06  <  0.11  :=  0.18
>=  0.11  <  0.16  :=  0.26
>=  0.16  <  0.21  :=  0.32
>=  0.21  <  0.26  :=  0.38
>=  0.26  <  0.31  :=  0.44
>=  0.31  <  0.36  :=  0.50
>=  0.36  <  0.41  :=  0.54
>=  0.41  <  0.46  :=  0.58
>=  0.46  <  0.51  :=  0.62
>=  0.51  <  0.56  :=  0.66
>=  0.56  <  0.61  :=  0.70
>=  0.61  <  0.66  :=  0.74
>=  0.66  <  0.71  :=  0.78
>=  0.71  <  0.76  :=  0.82
>=  0.76  <  0.81  :=  0.86
>=  0.81  <  0.86  :=  0.90
>=  0.86  <  0.91  :=  0.94
>=  0.91  <  0.96  :=  0.98
>=  0.96  <  1.01  :=  1.00



      PRINT FNpricefraction(0.5)
      END
 
      DEF FNpricefraction(p)
      IF p < 0.06 THEN = 0.10
      IF p < 0.11 THEN = 0.18
      IF p < 0.16 THEN = 0.26
      IF p < 0.21 THEN = 0.32
      IF p < 0.26 THEN = 0.38
      IF p < 0.31 THEN = 0.44
      IF p < 0.36 THEN = 0.50
      IF p < 0.41 THEN = 0.54
      IF p < 0.46 THEN = 0.58
      IF p < 0.51 THEN = 0.62
      IF p < 0.56 THEN = 0.66
      IF p < 0.61 THEN = 0.70
      IF p < 0.66 THEN = 0.74
      IF p < 0.71 THEN = 0.78
      IF p < 0.76 THEN = 0.82
      IF p < 0.81 THEN = 0.86
      IF p < 0.86 THEN = 0.90
      IF p < 0.91 THEN = 0.94
      IF p < 0.96 THEN = 0.98
      = 1.00