#include <bits/stdc++.h>
using namespace std;
unsigned long long int n, lo, hi, mid;
unsigned long long int sm(unsigned long long int p) {
  unsigned long long int s = 0;
  while (p > 0) {
    s += p % 10;
    p /= 10;
  }
  return s;
}
unsigned long long int power(unsigned long long int a,
                             unsigned long long int b) {
  if (b == 0) return 1;
  if (b & 1) return a * power(a, b - 1);
  unsigned long long int d = power(a, b >> 1);
  return d * d;
}
bool can(int x) {
  unsigned long long int y = (n - power(x, 2LL));
  if (x * sm(x) != y) return 0;
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  lo = 1;
  hi = floor(sqrt(n));
  if (hi * hi == n) hi--;
  while (hi - lo > 10000) {
    mid = lo + ((hi - lo) >> 1);
    if (can(mid))
      hi = mid;
    else
      lo = mid + 1;
  }
  for (; lo < hi; lo++)
    if (can(lo)) break;
  if (can(lo))
    cout << (lo) << endl;
  else
    cout << (-1) << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
unsigned long long int n, lo, hi, mid;
unsigned long long int sm(unsigned long long int p) {
  unsigned long long int s = 0;
  while (p > 0) {
    s += p % 10;
    p /= 10;
  }
  return s;
}
unsigned long long int power(unsigned long long int a,
                             unsigned long long int b) {
  if (b == 0) return 1;
  if (b & 1) return a * power(a, b - 1);
  unsigned long long int d = power(a, b >> 1);
  return d * d;
}
bool can(int x) {
  unsigned long long int y = (n - power(x, 2LL));
  if (x * sm(x) != y) return 0;
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  lo = 1;
  hi = floor(sqrt(n));
  if (hi * hi == n) hi--;
  while (hi - lo > 10000) {
    mid = lo + ((hi - lo) >> 1);
    if (can(mid))
      hi = mid;
    else
      lo = mid + 1;
  }
  for (; lo < hi; lo++)
    if (can(lo)) break;
  if (can(lo))
    cout << (lo) << endl;
  else
    cout << (-1) << endl;
  return 0;
}

```