The ionization enthalpy of Na⁺ formation from Na_(g) is 495.8 kJ mol^$-$1, while the electron gain enthalpy of Br is $-$325.0 kJ mol^$-$1. Given the lattice enthalpy of NaBr is $-$728.4 kJ mol^$-$1. The energy for the formation of NaBr ionic solid is ($-$) ____________ $\times$ 10^$-$1 kJ mol^$-$1.