Write code to complete DoublePennies()’s base case. Sample output for below program with inputs 1 and 10:

Number of pennies after 10 days: 1024

Note: If the submitted code has an infinite loop, the system will stop running the code after a few seconds, and report “Program end never reached.” The system doesn’t print the test case that caused the reported message

#include <stdio.h>

// Returns number of pennies if pennies are doubled numDays times
long long DoublePennies(long long numPennies, int numDays){
long long totalPennies;

/* Your solution goes here */

else {
totalPennies = DoublePennies((numPennies * 2), numDays – 1);
}

return totalPennies;
}

// Program computes pennies if you have 1 penny today,
// 2 pennies after one day, 4 after two days, and so on
int main(void) {
long long startingPennies;
int userDays;

scanf(“%lld”, &startingPennies);
scanf(“%d”, &userDays);
printf(“Number of pennies after %d days: %lld\n”, userDays, DoublePennies(startingPennies, userDays));

return 0;
}