#include <bits/stdc++.h>
using namespace std;
const unsigned long long INF = 1e10;
const unsigned long long N = 2e5 + 2;
const unsigned long long mod = 1e9 + 7;
const long double eps = 1E-7;
long double c, d, x1, x2, mn;
int s(int x) {
  if (x < 10) return x;
  return x % 10 + s(x / 10);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> c;
  mn = INF;
  for (long double b = 1; b <= 90; b++) {
    d = b * b + 4 * c;
    if (d < 0) continue;
    x1 = (sqrt(d) - b) / 2;
    x2 = (sqrt(d) + b) / 2;
    if (x1 > 0 && int(x1) == x1 && s(int(x1)) == b) mn = min(mn, x1);
    if (x2 > 0 && int(x2) == x2 && s(int(x2)) == b) mn = min(mn, x2);
  }
  if (mn == INF) mn = -1;
  cout << int(mn) << endl;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const unsigned long long INF = 1e10;
const unsigned long long N = 2e5 + 2;
const unsigned long long mod = 1e9 + 7;
const long double eps = 1E-7;
long double c, d, x1, x2, mn;
int s(int x) {
  if (x < 10) return x;
  return x % 10 + s(x / 10);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> c;
  mn = INF;
  for (long double b = 1; b <= 90; b++) {
    d = b * b + 4 * c;
    if (d < 0) continue;
    x1 = (sqrt(d) - b) / 2;
    x2 = (sqrt(d) + b) / 2;
    if (x1 > 0 && int(x1) == x1 && s(int(x1)) == b) mn = min(mn, x1);
    if (x2 > 0 && int(x2) == x2 && s(int(x2)) == b) mn = min(mn, x2);
  }
  if (mn == INF) mn = -1;
  cout << int(mn) << endl;
}

```