Problem: A positive integer a is given. Baron Munchausen claims that he knows such a positive integer n that if one multiplies n by a, the sum of its digits decreases a times. In other words, S(an) = S(n)/a, where S(x) denotes the sum of digits of the number x. 

Find out if what Baron told can be true.

Input

The only line contains a single integer a (2 ≤ a ≤ 10^3).

Output

If there is no such number n, print -1.

Otherwise print any appropriate positive integer n. Your number must not consist of more than 5⋅10^5 digits. We can show that under given constraints either there is no answer, or there is an answer no longer than 5⋅10^5 digits.

Examples

Input


2


Output


6


Input


3


Output


6669


Input


10


Output


-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int A;
const int MA = 1e3 + 10;
const int MC = 1e4 + 10;
const int MV = 2e6;
struct S {
 public:
  int c, b;
  S* p;
  int v;
  void setn(S& o, const int& x) {
    int w = c + A * x;
    o.c = w / 10, o.b = b + A * (w % 10) - x, o.p = this, o.v = x;
  }
  bool isd(void) const { return c == 0 && b == 0; }
  bool inb(void) const { return -A <= b && b <= A && c < MC; }
};
S s[MV];
int c;
using vb = bitset<MC>;
vb v[MC];
using qs = queue<int>;
qs q;
S* f;
void bfs(int k) {
  q.push(0);
  for (int i, u, x; !q.empty();) {
    i = q.front(), q.pop();
    S& n = s[i];
    for (u = 0; u < 10; u++) {
      if (!u && n.isd()) continue;
      n.setn(s[c], u);
      if (s[c].inb()) {
        if (s[c].isd()) {
          f = &s[c];
          return;
        }
        if (v[s[c].b + A][s[c].c]) continue;
        v[s[c].b + A][s[c].c] = 1;
        q.push(c++);
      }
    }
  }
}
void ans(S* n) {
  bool L = 0;
  for (; n->p; n = n->p)
    if (L || n->v) printf("%d", n->v), L = 1;
}
int main(void) {
  scanf("%d", &A);
  s[0] = {0, 0, nullptr, -1};
  c = 1;
  f = nullptr;
  bfs(0);
  if (f)
    ans(f), printf("\n");
  else
    printf("-1\n");
  return 0;
}