#include <bits/stdc++.h>
int main() {
  long long n, ans, minn = 100000000000, i;
  scanf("%I64d", &n);
  long long a, b, c, d, k, s, l = 0, m = 1, t, q, r, sum = 0;
  for (i = 1; i < 165; i++) {
    l = 0;
    m = 1;
    a = -i;
    b = i * i;
    c = 4 * n;
    d = b + c;
    if (d >= 0) {
      k = sqrt(d);
      if (d == k * k) l = 1;
    }
    if (l == 1) {
      a = (a + k);
      s = a / 2;
      if (a > 0 && (s * 2) == a) m = 0;
    }
    if (m == 0) {
      t = s;
      while (t != 0) {
        q = t % 10;
        sum = sum + q;
        t = t / 10;
      }
      if (sum == i) {
        ans = s;
        if (minn > s) minn = s;
      }
    }
    m = 1;
    if (l == 1) {
      a = (a - k);
      s = a / 2;
      if (a > 0 && (s * 2) == a) m = 0;
    }
    if (m == 0) {
      t = s;
      while (t != 0) {
        q = t % 10;
        sum = sum + q;
        t = t / 10;
      }
      if (sum == i) {
        ans = s;
        if (minn > s) minn = s;
      }
    }
  }
  if (minn > 0 && minn < 100000000000)
    printf("%I64d\n", minn);
  else
    printf("-1\n");
}

### Prompt

In CPP, your task is to solve 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>
int main() {
  long long n, ans, minn = 100000000000, i;
  scanf("%I64d", &n);
  long long a, b, c, d, k, s, l = 0, m = 1, t, q, r, sum = 0;
  for (i = 1; i < 165; i++) {
    l = 0;
    m = 1;
    a = -i;
    b = i * i;
    c = 4 * n;
    d = b + c;
    if (d >= 0) {
      k = sqrt(d);
      if (d == k * k) l = 1;
    }
    if (l == 1) {
      a = (a + k);
      s = a / 2;
      if (a > 0 && (s * 2) == a) m = 0;
    }
    if (m == 0) {
      t = s;
      while (t != 0) {
        q = t % 10;
        sum = sum + q;
        t = t / 10;
      }
      if (sum == i) {
        ans = s;
        if (minn > s) minn = s;
      }
    }
    m = 1;
    if (l == 1) {
      a = (a - k);
      s = a / 2;
      if (a > 0 && (s * 2) == a) m = 0;
    }
    if (m == 0) {
      t = s;
      while (t != 0) {
        q = t % 10;
        sum = sum + q;
        t = t / 10;
      }
      if (sum == i) {
        ans = s;
        if (minn > s) minn = s;
      }
    }
  }
  if (minn > 0 && minn < 100000000000)
    printf("%I64d\n", minn);
  else
    printf("-1\n");
}

```