Design a stack-like data structure to push elements to the stack and pop the most frequent element from the stack.

Implement the `FreqStack`

class:

`FreqStack()`

constructs an empty frequency stack.`void push(int val)`

pushes an integer`val`

onto the top of the stack.`int pop()`

removes and returns the most frequent element in the stack.- If there is a tie for the most frequent element, the element closest to the stack's top is removed and returned.

**Example 1:**

Input["FreqStack", "push", "push", "push", "push", "push", "push", "pop", "pop", "pop", "pop"] [[], [5], [7], [5], [7], [4], [5], [], [], [], []]Output[null, null, null, null, null, null, null, 5, 7, 5, 4]ExplanationFreqStack freqStack = new FreqStack(); freqStack.push(5); // The stack is [5] freqStack.push(7); // The stack is [5,7] freqStack.push(5); // The stack is [5,7,5] freqStack.push(7); // The stack is [5,7,5,7] freqStack.push(4); // The stack is [5,7,5,7,4] freqStack.push(5); // The stack is [5,7,5,7,4,5] freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,5,7,4]. freqStack.pop(); // return 7, as 5 and 7 is the most frequent, but 7 is closest to the top. The stack becomes [5,7,5,4]. freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,4]. freqStack.pop(); // return 4, as 4, 5 and 7 is the most frequent, but 4 is closest to the top. The stack becomes [5,7].

**Constraints:**

`0 <= val <= 10`

^{9}- At most
`2 * 10`

calls will be made to^{4}`push`

and`pop`

. - It is guaranteed that there will be at least one element in the stack before calling
`pop`

.