What will be the output of the program if it is executed like below?
cmd> sample

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

int main(int argc, char **argv)
{
    printf("%s\n", argv[argc-1]);
    return 0;
}

#include<stdio.h>

int main()
{
    printf("%x\n", -1<<3);
    return 0;
}

#include<stdio.h>
#include<string.h>

int main()
{
    printf("%c\n", "abcdefgh"[4]);
    return 0;
}

#include<stdio.h>
int main()
{
    union a
    {
        int i;
        char ch[2];
    };
    union a u;
    u.ch[0] = 3;
    u.ch[1] = 2;
    printf("%d, %d, %d\n", u.ch[0], u.ch[1], u.i);
    return 0;
}

#include<stdio.h>
int main()
{
    int X=40;
    {
        int X=20;
        printf("%d ", X);
    }
    printf("%d\n", X);
    return 0;
}

#include<stdio.h>

int main()
{
    char c=48;
    int i, mask=01;
    for(i=1; i<=5; i++)
    {
        printf("%c", c|mask);
        mask = mask<<1;
    }
    return 0;
}

#include<stdio.h>

int main()
{
    char str = "CuriousTab";
    printf("%s\n", str);
    return 0;
}

#include<stdio.h>

int main()
{
    static char s[25] = "The cocaine man";
    int i=0;
    char ch;
    ch = s[++i];
    printf("%c", ch);
    ch = s[i++];
    printf("%c", ch);
    ch = i++[s];
    printf("%c", ch);
    ch = ++i[s];
    printf("%c", ch);
    return 0;
}

#include<stdio.h>

int main()
{
    printf("%d %d\n", 32<<1, 32<<0);
    printf("%d %d\n", 32<<-1, 32<<-0);
    printf("%d %d\n", 32>>1, 32>>0);
    printf("%d %d\n", 32>>-1, 32>>-0);
    return 0;
}

#include<stdio.h>

int main()
{
    printf("%d >> %d %d >> %d\n", 4 >> 1, 8 >> 1);
    return 0;
}