Competitive Programming #13 : [The Mighty Divisible Number]

Given four numbers X, Y, Z and N. Your task is to find smallest N digit Number that is divisible by all the three numbers X, Y and Z.
Note: If no such number can be formed then print "-1".
Input:
The first line of the input contains an integer T, denoting the number of test cases. Then T test case follows. The only line of each test case contains four space separated integers denoting the values of X, Y, Z and N respectively.

Output:
For each test case output the smallest N digit number divisible by X, Y and Z. If no such number can be formed then print "-1".

Constraints:
1<=T<=10³
1<=X,Y,Z<=10³
1<=N<=18

Example:
Input:
3
2 3 5 4
4 5 6 3
3 5 7 2
Output:
1020
120
-1

Solution:
 
So how gonna this works ?
The most important thing here is LCM (x,y,z);
if LCM is greater or equal ten*10 it is impossible to get result about any three numbers (ten is smallest n digits number, example: the smallest 3 digits number is 1000 ) 
if LCM == 1 (if  lcm*ten = ten) then  print value of ten;
if LCM is smaller than ten then printf ten + (LCM-(ten%LCM)); 
else LCM is more than ten && LCM is smaller than ten*10 then  printf LCM;