# Even Length Palindromic Number

## Problem:-

You have to design a new model which maps an even length palindromic number to some digit between 0 to 9.
The number is mapped to a digit

$x$

on the basis of following criteria:
1.

$x$

should appear maximum number of times in the palindromic number, that is, among all digits in the number,

$x$

should appear maximum number of times.
2. If more than one digit appears maximum number of times,

$x$

should be the smallest digit among them.

Given an integer

$N$

, you have to find the digit

$x$

for the

${N}^{th}$

even length palindromic number.
Note- First 9 even length palindromic numbers are:
11, 22, 33, 44, 55, 66, 77, 88, 99
Input :
First line contains T, number of test cases.
Each of the next T lines contains an integer N.
Output:
For each test case, print the digit to which the

${N}^{th}$

even length palindromic number is mapped.
Answer for each test case should come in a new line.
Constraints:

$1\le T\le {10}^{5}$

$1\le N\le {10}^{18}$

SAMPLE INPUT

`31210`
SAMPLE OUTPUT

`120`
Explanation

For case 1:
1st even length palidromic number is 11 , so answer is 1 as 1 appears most number of times in the number.
For case 2:
2nd even length palidromic number is 22 , so answer is 2 as 2 appears most number of times in the number.
For case 3:
10th even length palindromic number is 1001, here both 0 and 1 appears same number of times but 0<1 so answer is 0.
Time Limit:1.0 sec(s) for each input file.
Memory Limit:256 MB
Source Limit:1024 KB

## solution:-

#include<stdio.h>
void main()
{
int t,k=0,a,max,r;
long long int n;
scanf(“%d”,&t);
for(int i=0;i<t;i++)
{
for(int j=0;j<=9;j++)
a[j]=0;
scanf(“%lld”,&n);
while(n!=0)
{
r=n%10;
a[r]++;
n=n/10;
}
max=a;
for(int j=0;j<=9;j++)
{
if(max<a[j])
{
max=a[j];
k=j;
}
}
printf(“%dn”,k);
k=0;
}
}