If 1+(2+^49 C_1+^49 C_2+…+^49 C_49)(^50 C_2+^50 C_4+…+^50 C_50) is equal to 2^n · m, where m is odd, then n + m is equal to \ \ \ \.

If $\displaystyle 1+\left(2+{ }^{49} \mathrm{C}_1+{ }^{49} \mathrm{C}_2+\ldots+{ }^{49} \mathrm{C}_{49}\right)\left({ }^{50} \mathrm{C}_2+{ }^{50} \mathrm{C}_4+\ldots+{ }^{50} \mathrm{C}_{50}\right)$ is equal to $\displaystyle 2^{\mathrm{n}} \cdot \mathrm{m}$, where m is odd, then $\displaystyle \mathrm{n}+\mathrm{m}$ is equal to \_\_\_\_.