Non-square Equation CodeForces - 233B （数学）

Let’s consider equation:

x2 + s(x)·x - n = 0, 

where x, n are positive integers, s(x) is the function, equal to the sum of digits of number x in the decimal number system.

You are given an integer n, find the smallest positive integer root of equation x, or else determine that there are no such roots.

Input

A single line contains integer n (1 ≤ n ≤ 10^18) — the equation parameter.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Output

Print -1, if the equation doesn’t have integer positive roots. Otherwise print such smallest integer x (x > 0), that the equation given in the statement holds.

Example

Input

2

Output

1

Input

110

Output

10

Input

4

Output

-1

Note

In the first test case x = 1 is the minimum root. As s(1) = 1 and 12 + 1·1 - 2 = 0.

In the second test case x = 10 is the minimum root. As s(10) = 1 + 0 = 1 and 102 + 1·10 - 110 = 0.

In the third test case the equation has no roots.

思路：假如x有解，那么x < sqrt(n)，所以x的范围小于1e9，也就是说x的长度最大不超过9，考虑每个数位上的最大值不超过9，所以s(x)范围为1到81，那么我们枚举s（x）的值，然后解下x的一元二次方程，然后判断下所求得解是否满足s（x）。

代码如下

#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long
using namespace std;
int judeg(LL x,int sum)
{
int s=0;
while(x>0)
{
s+=x%10;
x/=10;
}
if(s==sum) return 1;
else return 0;
}
LL min(LL a,LL b)
{
if(a<b)
return a;
return b;
}
int main()
{
LL n;
cin>>n;
int ans=1e9+5;
for(int i=1;i<=81;i++)
{
double a=sqrt(n+i*i/4)-i/2;
LL b=(LL)a;
if(b*b+b*i-n==0&&judeg(b,i))
ans=min(ans,b);
}
if(ans!=1e9+5)
printf("%d\n",ans);
else
printf("-1\n");
return 0;
}