how to repeat a menu after dealing with wrong input - c

Question

So the assignemt is to print a menu with options, and if the user puts in an invalid choice (not 1,2,3,4,5 ,6), it prints an error and asking the user to choose again.
if the user puts in a wrong input 5 times total, the programm will exit.
int main() {
printf("Welcome, please choose one of the options below: \n ");
printf( "1.Exit \n ");
printf( "2.Print menu again \n ");
printf( "3. ");
printf( "4.. ");
printf( "5. ");
printf( "6. ");
printf("Enter your choice: ");
scanf("%d" , &choice);
if( (choice > 6) || (choice < 1) ) {
do {
count++;
printf(" Wrong input, please try again (Enter 2 for re-printing the menu). \n " );
printf("Enter your choice: ");
scanf("%d", &choice);
if(choice==2){
do {
printf("Welcome, please choose one of the options below: \n "); //prints of the screen the following in a loop
printf("1.Exit \n ");
printf("2.Print menu again \n ");
printf("3. ");
printf("4. ");
printf("5. ");
printf("6.");
printf("Enter your choice: ");
scanf("%d", &choice);
} while (choice==2);
}
} while (count < 4) ;
printf("%s" , "You have made 5 menu errors.Bye Bye!!! \n ");
}
while(1) {
.
}
*the while(1)is for the entire code, puts the entire code for re-use
** i didnt use switch-case, cuz it's forbidden to use it
Now, the problem is, that if i put a wrong input first, let's say for example, '7' (which isn't a choice from the menu), it will print "wrong input, please try again". So far so good.
But then,if I press 2 for re-printing the menu, and then I press any number, even if it's a valid choice, it's printing "wrong input".
also, if I press '2' for re-printing the menu, and then press 1, it will require to press 1 twice in order to exit the programm, instead of just pressing once.

Answer 1

The above answers look correct but you can go with the following code as its working and easy to understand for anyone!
#include <stdio.h>
void printMenu()
{
printf("Welcome, please choose one of the options below: \n ");
printf( "1.Exit \n ");
printf( "2.Print menu again \n ");
printf( "3. ");
printf( "4.. ");
printf( "5. ");
printf( "6. ");
}
int main()
{
int choiceValid=0, count=0, choice;
printMenu();
while(choiceValid==0 && count<=5)
{
printf("Enter your choice: ");
scanf("%d" , &choice);
if(choice==2)
{
printMenu();
continue;
}
if( choice<=6 && choice>=1 )
choiceValid=1;
else
{
count++;
printf("\nWrong input, please try again (Enter 2 for re-printing the menu). \n " );
}
}
return 0;
}

Answer 2

Replace your if(choice==2) with if((choice==2) || (choice > 6) || (choice < 1) )
Replace while (choice==2);with while (((choice==2) || (choice > 6) || (choice < 1) ) && (count < 4))
Put the following block inside the same while loop.
if(2 == choice) count = 0;
else if (choice > 6) || (choice < 1) count ++;