CodeForces

B. Non-square Equation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ≤ 10¹⁸) — 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.

Examples

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.

枚举sx sx的范围可以看出最多不会超过100
由sx求x公式

#include <bits/stdc++.h>using namespace std;#define LL long longLL get(LL n) {    LL sum=0;    while(n) {        sum += n%10;        n /= 10;    }    return sum;}int main() {    LL n, ans = -1;    scanf("%lld",&n);    for(int i = 1;i < 100; i++) {        LL x = sqrt(i*i/4+n) - i/2;        LL sx = get(x);        if(x*x+sx*x==n) {            ans = x;            break;        }    }    printf("%d\n",ans); }