int main(void) {         int a;         char *s;         int v0 = 9, v1 = 0, v2 = 0, v3 = 0;           printf("Exercise 1:\n====

Run the following code and answer the questions.

#include <stdio.h>

int main(void) {

        int a;

        char *s;

        int v0 = 9, v1 = 0, v2 = 0, v3 = 0;

        printf("Exercise 1:\n====================\n");

        switch(v0) {

               case 0: printf("Good day\n");   break;

               case 1: printf("Hey There!\n");    break;

               case 2: printf("No Name\n");  break;

               case 3: printf("Yes Please\n"); break;

               case 4: printf("Go ");        

               case 5: printf("Vamos!\n");                

               default: printf(" ");               break;

        }

        for(v1; v1<v2; v1=v1+v1) {

               printf("Vamos ");

        }

        printf("\n");

        if (v2 == 6) {

            s = "Go";

        }

        if(v3 != 0) {

               printf("%s VAMOS!\n",s);

        }

        return 0;

}

1. What is the output of the program? Please explain why using the variables v0, v1, v2, and v3.
2. The code has a vulnerability that can prevent the program from outputting properly. Explain what the problem is. What are the values of v0, v1, v2, and v3 that may cause such problem?