#include <bits/stdc++.h>
using namespace std;
long long bsearch(long long n) {
  long long lo = 1, hi = 4000000000, mid;
  for (int i = 0; i < 500; i++) {
    mid = (lo + hi) / 2;
    if (mid * mid == n)
      return mid;
    else {
      if (mid * mid < n)
        lo = mid;
      else
        hi = mid;
    }
  }
  return -1;
}
long long sumdig(long long n) {
  long long sum = 0;
  while (n > 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  long long n, ans = 1000000000, a;
  cin >> n;
  for (long long s = 1; s <= 100; s++) {
    a = bsearch(s * s + 4 * n);
    if (a == -1) continue;
    if ((a - s) % 2 != 0) continue;
    if (sumdig((a - s) / 2) != s) continue;
    if ((a - s) / 2 <= 0) continue;
    ans = min(ans, (a - s) / 2);
  }
  if (ans == 1000000000)
    cout << "-1\n";
  else
    cout << ans << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 bsearch(long long n) {
  long long lo = 1, hi = 4000000000, mid;
  for (int i = 0; i < 500; i++) {
    mid = (lo + hi) / 2;
    if (mid * mid == n)
      return mid;
    else {
      if (mid * mid < n)
        lo = mid;
      else
        hi = mid;
    }
  }
  return -1;
}
long long sumdig(long long n) {
  long long sum = 0;
  while (n > 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  long long n, ans = 1000000000, a;
  cin >> n;
  for (long long s = 1; s <= 100; s++) {
    a = bsearch(s * s + 4 * n);
    if (a == -1) continue;
    if ((a - s) % 2 != 0) continue;
    if (sumdig((a - s) / 2) != s) continue;
    if ((a - s) / 2 <= 0) continue;
    ans = min(ans, (a - s) / 2);
  }
  if (ans == 1000000000)
    cout << "-1\n";
  else
    cout << ans << "\n";
  return 0;
}

```