Create a new stack in which to put every third element of the first stack

Hello!
I need some help.
I have to do the following task:
Create a stack with numbers in the range from -50 to +50. After creating the stack, perform the following actions: create a new stack, in which every third element of the first stack will be placed. Upon completion, all stacks must be removed.

I was able to write the code of the program with creating a stack in the specified range, viewing and deleting it, but I can not implement the function to create the second stack.

Code:

---

#include "stdafx.h"
#include <iomanip>
#include <stack>
#include <iostream>
#include <ctime>
using namespace std;

struct Stack   
{
        int info;
        int in;
        Stack *next;
} *begin1, *second; 

Stack* InStack(Stack*, int);
void View(Stack*);
Stack* DelStackAll(Stack*);

int main()
{        srand(time(0));
        int i; 
        for(i=1; i<=10; i++)
        {
                begin1=InStack(begin1, rand()%101-50);
        }
        View(begin1);
       
        begin1=DelStackAll(begin1);
       
        getchar();
}

Stack* InStack(Stack *p, int in){
        Stack *t=new Stack;
        t->info=in;       
        t->next=p;         
        return t;
}

void View(Stack *p)   
{
        Stack *t=p;     
        while (t!=NULL)
        {
                cout<<setw(5)<<t->info; 
                t=t->next;           
        }
        cout<<endl;
}

Stack* DelStackAll(Stack *p)
{ Stack *t;
        while(p != NULL) {
                t=p;
                p=p->next;
                delete t;
        }
        return p;
}

---

I will be very grateful for the help!

- you need to first initialise begin1 and second to nullptr as currently the first time begin1 is passed to InStack it's value is undefined.

- iterate the stack begin1 and for every third entry (keep a counter of the number of entries in the iteration and if % 3 (mod 3) is 0 then you have a third entry) then just call InStack(second, n) where n is the value from the iteration of begin1. This will create the stack second. Then just use View(second) to display the second stack etc.

Ok, thank you,I understood what I need to do, but there may be another way to do this operation without a counter, since I need to write this function without using what you described.