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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def _Melting_Pressure(T, ice="Ih"):
 """Melting Pressure correlation Parameters T : float Temperature, [K] ice: string Type of ice: Ih, III, V, VI, VII. Below 273.15 is a mandatory input, the ice Ih is the default value. Above 273.15, the ice type is unnecesary. Returns ------- P : float Pressure at sublimation line, [MPa] Notes ------ Raise :class:`NotImplementedError` if input isn't in limit: * 251.165 ≤ T ≤ 715 Examples -------- 8.947352740189152e-06 268.6846466336108 References IAPWS, Revised Release on the Pressure along the Melting and Sublimation Curves of Ordinary Water Substance, http://iapws.org/relguide/MeltSub.html. """

    if ice == "Ih" and 251.165 <= T <= 273.16:
        # Ice Ih
        Tref = Tt
        Pref = Pt
        Tita = T/Tref
        a = [0.119539337e7, 0.808183159e5, 0.33382686e4]
        expo = [3., 0.2575e2, 0.10375e3]
        suma = 1
        for ai, expi in zip(a, expo):
            suma += ai*(1-Tita**expi)
        P = suma*Pref
    elif ice == "III" and 251.165 < T <= 256.164:
        # Ice III
        Tref = 251.165
        Pref = 208.566
        Tita = T/Tref
        P = Pref*(1-0.299948*(1-Tita**60.))
    elif (ice == "V" and 256.164 < T <= 273.15) or 273.15 < T <= 273.31:
        # Ice V
        Tref = 256.164
        Pref = 350.100
        Tita = T/Tref
        P = Pref*(1-1.18721*(1-Tita**8.))
    elif 273.31 < T <= 355:
        # Ice VI
        Tref = 273.31
        Pref = 632.400
        Tita = T/Tref
        P = Pref*(1-1.07476*(1-Tita**4.6))
    elif 355. < T <= 715:
        # Ice VII
        Tref = 355
        Pref = 2216.000
        Tita = T/Tref
        P = Pref*exp(1.73683*(1-1./Tita)-0.544606e-1*(1-Tita**5) +
                     0.806106e-7*(1-Tita**22))
    else:
        raise NotImplementedError("Incoming out of bound")
    return P