Problem:
Score : 500 points

Problem Statement

For integers b (b \geq 2) and n (n \geq 1), let the function f(b,n) be defined as follows:

f(b,n) = n, when n < b

f(b,n) = f(b,\,{\rm floor}(n / b)) + (n \ {\rm mod} \ b), when n \geq b

Here, {\rm floor}(n / b) denotes the largest integer not exceeding n / b,
and n \ {\rm mod} \ b denotes the remainder of n divided by b.

Less formally, f(b,n) is equal to the sum of the digits of n written in base b.
For example, the following hold:

f(10,\,87654)=8+7+6+5+4=30

f(100,\,87654)=8+76+54=138

You are given integers n and s.
Determine if there exists an integer b (b \geq 2) such that f(b,n)=s.
If the answer is positive, also find the smallest such b.

Constraints

1 \leq n \leq 10^{11}

1 \leq s \leq 10^{11}

n,\,s are integers.

Input

The input is given from Standard Input in the following format:

n
s

Output

If there exists an integer b (b \geq 2) such that f(b,n)=s, print the smallest such b.
If such b does not exist, print -1 instead.

Sample Input 1

87654
30

Sample Output 1

10

Sample Input 2

87654
138

Sample Output 2

100

Sample Input 3

87654
45678

Sample Output 3

-1

Sample Input 4

31415926535
1

Sample Output 4

31415926535

Sample Input 5

1
31415926535

Sample Output 5

-1

Solution in Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var
	n,s,b,t,c,ans,i:int64;
function f(b,n:int64):int64;
var t,c:int64;
begin
	t:=n;
	c:=0;
	while t>0 do begin
		c:=c+t mod b;
		t:=t div b;
	end;
	f:=c;
end;
begin
	read(n,s);
	i:=1;
	ans:=n+b;
	while i*i<=n do begin
		if(i>1)and(f(i,n)=s)and(ans>i)then ans:=i;
		if n div i>1 then begin
			t:=f(n div i,n);
			if(t<=s)or((s-t)mod i=0)then begin
				b:=n div i+(s-t)div i;
				if(b>1)and(f(b,n)=s)and(ans>b)then ans:=b;
			end;
		end;
		inc(i);
	end;
	if ans=n+b then writeln(-1)else writeln(ans);
end.