What will be the output of the program (myprog.c) given below if it is executed from the command line?
cmd> myprog 1 2 3

/* myprog.c */
#include<stdio.h>
#include<stdlib.h>

int main(int argc, char **argv)
{
    int i, j=0;
    for(i=0; i<argc; i++)
        j = j+atoi(argv[i]);
    printf("%d\n", j);
    return 0;
}

/* myprog.c */
#include<stdio.h>

int main(int argc, char *argv[])
{
    int i;
    for(i=1; i<argc; i++)
        printf("%c", argv[i][0]);
    return 0;
}

#include<stdio.h>

int main(int argc, char *argv, char *env[])
{
    int i;
    for(i=1; i<argc; i++)
        printf("%s\n", env[i]);
    return 0;
}

/* myproc.c */
#include<stdio.h>

int main(int argc, char *argv[])
{
    printf("%s", argv[0]);
    return 0;
}

/* sample.c */
#include<stdio.h>

int main(int argc, char *argv[])
{
    int i=0;
    i+=strlen(argv[1]);
    while(i>0)
    {
        printf("%c", argv[1][--i]);
    }
    return 0;
}

/* sample.c */
#include<stdio.h>

int main(int argc, char *argv[])
{
    printf("%c", **++argv);
    return 0;
}

#include<stdio.h>
void fun(int);

int main(int argc)
{
    printf("%d ", argc);
    fun(argc);
    return 0;
}
void fun(int i)
{
    if(i!=4)
        main(++i);
}

/* myprog.c */
#include<stdio.h>

int main(int argc, char *argv[])
{
    printf("%c", *++argv[1]);
    return 0;
}

/* myprog.c */
#include<stdio.h>

int main(int argc, char **argv)
{
    printf("%c\n", **++argv);
    return 0;
}

/* sample.c */
#include<stdio.h>

int main(int argc, char *argv[])
{
    int j;
    j = argv[1] + argv[2] + argv[3];
    printf("%d", j);
    return 0;
}

/* sample.c */
#include<stdio.h>

int main(int argc, char *argv[])
{
    printf("%s\n", argv[0]);
    return 0;
}