Comparing Two Long Integers

Description

You are given two very long integers a, b (leading zeroes are allowed). You should check what numbera or b is greater or determine that they are equal.

The input size is very large so don't use the reading of symbols one by one. Instead of that use the reading of a whole line or token.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to usescanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead ofScanner/System.out in Java. Don't use the function input() inPython2 instead of it use the function raw_input().

Input

The first line contains a non-negative integer a.

The second line contains a non-negative integer b.

The numbers a, b may contain leading zeroes. Each of them contains no more than10⁶ digits.

Output

Print the symbol "<" if a < b and the symbol ">" ifa > b. If the numbers are equal print the symbol "=".

Sample Input

Input

910

Output

<

Input

1110

Output

>

Input

0001234512345

Output

=

Input

01239

Output

>

Input

0123111

Output

>

#include <iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#include<cstdlib>using namespace std;int main(){    char a[1000005], b[1000005];    int n, m, i, j, x = 0, y = 0;    scanf("%s%s", a, b);    n = strlen(a);    m=strlen(b);    for (i = 0; i < n; i++)    {         if (a[i] == '0')            x++;         else            break;    }    for (i = 0; i < m; i++)    {         if (b[i] == '0')            y++;         else            break;    }    if (n - x > m - y)         printf(">\n");    else if (n - x < m - y)        printf("<\n");    else    {        for (i = x; i < n; i++)        {            if (a[i] > b[y])            {                printf(">\n");                break;            }            else if (a[i] < b[y])            {                 printf("<\n");                 break;            }            else if (a[i] == b[y])            {                y++;                continue;            }        }         if (i == n)            printf("=\n");    }    return 0;}