Problem:
Score : 100 points

Problem Statement

You are given a grid with 2 rows and 3 columns of squares.
The color of the square at the i-th row and j-th column is represented by the character C_{ij}.

Write a program that prints YES if this grid remains the same when rotated 180 degrees, and prints NO otherwise.

Constraints

C_{i,j}(1 \leq i \leq 2, 1 \leq j \leq 3) is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

C_{11}C_{12}C_{13}
C_{21}C_{22}C_{23}

Output

Print YES if this grid remains the same when rotated 180 degrees; print NO otherwise.

Sample Input 1

pot
top

Sample Output 1

YES

This grid remains the same when rotated 180 degrees.

Sample Input 2

tab
bet

Sample Output 2

NO

This grid does not remain the same when rotated 180 degrees.

Sample Input 3

eye
eel

Sample Output 3

NO

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>
#include <string.h>
/*int main()
{
    long long a[3][1000000010];
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<3;i++)
        {
            for(int j=0;j<n;j++)
            {
                scanf("%lld",&a[i][j]);
            }
        }
    }
}*/
/*int main()
{
    int k;
    int sum;
    while(~scanf("%d",&k))
    {
        if(k<10)
            sum=k;
        if(k>=10&&k<100)
        {
            int l1=k%10;
            int l2=k/10;
            sum=l1+l2;
        }
        if(k>=100&&k<1000)
        {
            int l1=k%10;
            int l2=k/10%10;
            int l3=k/100;
            sum=l1+l2+l3;
        }
        if(k>=1000&&k<10000)
        {
            int l1=k%10;
            int l2=k/10%10;
            int l3=k/100%10;
            int l4=k/1000;
            sum=l1+l2+l3+l4;
        }
        if(k>=10000&&k<100000)
        {
            int l1=k%10;
            int l2=k/10%10;
            int l3=k/100%10;
            int l4=k/1000%10;
            int l5=k/10000;
            sum=l1+l2+l3+l4+l5;
        }
        printf("%d\n",sum);
    }
    return 0;
}*/
char a[5][5];
int main()
{
    while(1)
    {
        for(int i=0; i<2; i++)
        {
            scanf("%s",&a[i]);
        }
        if(a[0][0]==a[1][2]&&a[0][1]==a[1][1]&&a[0][2]==a[1][0])
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    return 0;
}