Alternative Scale of Notation
Time Limit: 1 Second Memory Limit: 32768 KB
One may define a map of strings over an alphabet 
= { C1, C2, . . . CB } of size B to non-negative integer
numbers, using characters as digits C1 = 0, C2 = 1, . . . , CB = B - 1 and interpreting the string as the representation of some number in a scale of notation with base B. Let us denote this map by UB , for a string 
[ 1...n ] of length n we put
For example, U3(1001) = 1*27 + 0*9 + 0*3 + 1*1 = 28.
However, this correspondence has one major drawback: it is not ont-to-one. For example,
In mathematical logic and computer science this may be unacceptable. To overcome this problem, the
alternative interpretation is used. Let us interpret characters as digits, but in a slightly different way:
C1 = 1, C2 = 2, . . . , CB = B . Note that now we do not have 0 digit, but rather we have a rudiment B
digit. Now we define the map VB in a similar way, for each string 
[ 1...n ] of length n we put
For an empty string 
we put VB( 
) = 0.
This map looks very much like UB , however, the set of digits is now different. So, for example, we have V3(1313) = 1*27 + 3*9 + 1*3 + 3*1 = 60.
It can be easily proved that the correspondence defined by this map is one-to-one and onto. Such a map
is called bijective, and it is well known that every bijective map has an inverse. Your task in this problem
is to compute the inverse for the map VB . That is, for a given integer number x you have to find the
string 
, such that VB( 
) = x.
Input
Input consists of several test cases. For each case, the first line contains B (2 <= B <= 9) and the second line contains an integer number x given in a usual decimal scale of notation, 0 <= x <= 10100.
Output
For each test case, output in one line such string 
, consisting only of digits from the set { 1, 2, . . . , B }, that VB( 
) = x .
Sample Input
3 60 3 0 2 31
Sample Output
1313 11111Submit
Source: Northeastern Europe 2003