## Algorithm

A. Boring Apartments
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

There is a building consisting of 10 00010 000 apartments numbered from 11 to 10 00010 000, inclusive.

Call an apartment boring, if its number consists of the same digit. Examples of boring apartments are 11,2,777,999911,2,777,9999 and so on.

Our character is a troublemaker, and he calls the intercoms of all boring apartments, till someone answers the call, in the following order:

• First he calls all apartments consisting of digit 11, in increasing order (1,11,111,11111,11,111,1111).
• Next he calls all apartments consisting of digit 22, in increasing order (2,22,222,22222,22,222,2222)
• And so on.

The resident of the boring apartment x answers the call, and our character stops calling anyone further.

Our character wants to know how many digits he pressed in total and your task is to help him to count the total number of keypresses.

For example, if the resident of boring apartment 2222 answered, then our character called apartments with numbers 1,11,111,1111,2,221,11,111,1111,2,22 and the total number of digits he pressed is 1+2+3+4+1+2=131+2+3+4+1+2=13.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1t361≤�≤36) — the number of test cases.

The only line of the test case contains one integer x (1x99991≤�≤9999) — the apartment number of the resident who answered the call. It is guaranteed that x consists of the same digit.

Output

For each test case, print the answer: how many digits our character pressed in total.

Example
input
Copy
4
22
9999
1
777

output
Copy
13
90
1
66


## Code Examples

### #1 Code Example with C++ Programming

Code - C++ Programming

#include<bits/stdc++.h>
using namespace std;

int main(){
int t;
cin>>t;
string n;
int count = 0;
while(t--){
cin>>n;
count = 0;
count += (stoi(n)%10 - 1) * 10;
if(n.size() == 1){
count += 1;
cout<<count<<endl;
continue;
}
if(n.size() == 2){
count += 3;
cout<<count<<endl;
continue;
}
if(n.size() == 3){
count += 6;
cout<<count<<endl;
continue;
}
if(n.size() == 4){
count += 10;
cout<<count<<endl;
continue;
}
}
return 0;
}
Copy The Code &

Input

cmd
4
22
9999
1
777

Output

cmd
13 90
1
66