Set Nth Bit

Given a non-negative integer `num` and a 1-indexed bit position `n`, your task is to return the new number after setting the `n`-th bit of `num` to `1`. If the `n`-th bit is already `1`, it should remain `1`. All other bits should remain unchanged. The bits are 1-indexed from the rightmost (least significant) bit. For example, `n=1` refers to the rightmost bit, `n=2` refers to the second bit from the right, and so on. ### Examples: **Example 1:** Input: `num = 13`, `n = 2` Output: `15` Explanation: Binary of 13: `...01101` The 2nd bit (from the right) is `0`. Setting the 2nd bit to `1` results in `...01111`, which is 15 in decimal. **Example 2:** Input: `num = 7`, `n = 3` Output: `7` Explanation: Binary of 7: `...00111` The 3rd bit (from the right) is already `1`. Setting it to `1` again leaves the number unchanged. **Example 3:** Input: `num = 0`, `n = 1` Output: `1` Explanation: Binary of 0: `...00000` Setting the 1st bit to `1` results in `...00001`, which is 1 in decimal. ### Constraints: * `0 <= num <= 2 * 10^9` * `1 <= n <= 31`