#include <bits/stdc++.h>
using namespace std;
long long n;
int sum_d(int x) {
  int sum = 0;
  while (x > 0) {
    sum += x % 10;
    x /= 10;
  }
  return sum;
}
long long f(long long x, long long sumd) { return x * (x + sumd) - n; }
long long search(long long sumd) {
  long long lo = 1, hi = 1e9;
  while (hi - lo > 0) {
    long long mi = lo + (hi - lo) / 2;
    if (f(mi, sumd) >= 0)
      hi = mi;
    else
      lo = mi + 1;
  }
  if (f(lo, sumd) == 0) return lo;
  return -1;
}
int main() {
  cin >> n;
  long long mini = 1e9;
  for (int sumd = 1; sumd <= 100; sumd++) {
    long long x = search(sumd);
    if (x != -1 && sumd == sum_d(x)) mini = min(mini, x);
  }
  if (mini == 1e9)
    cout << "-1";
  else
    cout << mini;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
long long n;
int sum_d(int x) {
  int sum = 0;
  while (x > 0) {
    sum += x % 10;
    x /= 10;
  }
  return sum;
}
long long f(long long x, long long sumd) { return x * (x + sumd) - n; }
long long search(long long sumd) {
  long long lo = 1, hi = 1e9;
  while (hi - lo > 0) {
    long long mi = lo + (hi - lo) / 2;
    if (f(mi, sumd) >= 0)
      hi = mi;
    else
      lo = mi + 1;
  }
  if (f(lo, sumd) == 0) return lo;
  return -1;
}
int main() {
  cin >> n;
  long long mini = 1e9;
  for (int sumd = 1; sumd <= 100; sumd++) {
    long long x = search(sumd);
    if (x != -1 && sumd == sum_d(x)) mini = min(mini, x);
  }
  if (mini == 1e9)
    cout << "-1";
  else
    cout << mini;
  return 0;
}

```