Design a Stack With Increment Operation

Design a stack that supports increment operations on its elements. Implement the `CustomStack` class: - `CustomStack(int maxSize)` Initializes the object with `maxSize` which is the maximum number of elements in the stack. - `void push(int x)` Adds `x` to the top of the stack if the stack has not reached the `maxSize`. - `int pop()` Pops and returns the top of the stack or `-1` if the stack is empty. - `void inc(int k, int val)` Increments the bottom `k` elements of the stack by `val`. If there are less than `k` elements in the stack, increment all the elements in the stack.   **Example 1:** **Input** ["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"] [[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]] **Output** [null,null,null,2,null,null,null,null,null,103,202,201,-1] **Explanation** CustomStack stk = new CustomStack(3); // Stack is Empty [] stk.push(1); // stack becomes [1] stk.push(2); // stack becomes [1, 2] stk.pop(); // return 2 --> Return top of the stack 2, stack becomes [1] stk.push(2); // stack becomes [1, 2] stk.push(3); // stack becomes [1, 2, 3] stk.push(4); // stack still [1, 2, 3], Do not add another elements as size is 4 stk.increment(5, 100); // stack becomes [101, 102, 103] stk.increment(2, 100); // stack becomes [201, 202, 103] stk.pop(); // return 103 --> Return top of the stack 103, stack becomes [201, 202] stk.pop(); // return 202 --> Return top of the stack 202, stack becomes [201] stk.pop(); // return 201 --> Return top of the stack 201, stack becomes [] stk.pop(); // return -1 --> Stack is empty return -1. ```   **Constraints:** - `1 <= maxSize, x, k <= 1000` - `0 <= val <= 100` - At most `1000` calls will be made to each method of `increment`, `push` and `pop` each separately.