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
#include <bits/stdc++.h>
using namespace std;
template <class T>
int chkmax(T& a, T b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
int chkmin(T& a, T b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class iterator>
void output(iterator begin, iterator end, ostream& out = cerr) {
  while (begin != end) {
    out << (*begin) << " ";
    begin++;
  }
  out << endl;
}
template <class T>
void output(T x, ostream& out = cerr) {
  output(x.begin(), x.end(), out);
}
void fast_io() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int mx = 2000;
const int add = mx / 2;
const int D = 10;
int a, used[mx][mx], digit[mx][mx];
pair<int, int> mv[mx][mx];
void bfs() {
  queue<pair<int, int> > q;
  q.push({0, 0});
  while (!q.empty()) {
    int carry = q.front().first, balance = q.front().second;
    q.pop();
    int ok = 0;
    for (int x = 0; x < D; ++x) {
      if (carry == 0 && balance == 0 && x == 0) {
        continue;
      }
      int c2 = (carry + a * x) / D;
      int b2 = (balance + a * ((carry + a * x) % D) - x);
      int pos = b2 + add;
      if (pos >= 0 && pos < mx && c2 < mx && used[c2][pos] == 0) {
        digit[c2][pos] = x;
        used[c2][pos] = 1;
        q.push({c2, b2});
        mv[c2][pos] = {carry, balance + add};
        if (c2 == 0 && b2 == 0) {
          ok = 1;
          break;
        }
      }
    }
    if (ok) {
      break;
    }
  }
  if (used[0][add] == 0) {
    cout << -1 << endl;
  } else {
    int carry = 0, pos = add;
    vector<int> ans;
    while (true) {
      ans.push_back(digit[carry][pos]);
      int _carry = mv[carry][pos].first, _pos = mv[carry][pos].second;
      carry = _carry, pos = _pos;
      if (carry == 0 && pos == add) {
        break;
      }
    }
    int pos0 = 0;
    while (ans[pos0] == 0) {
      pos0++;
    }
    for (int i = pos0; i < ans.size(); ++i) {
      cout << ans[i];
    }
    cout << endl;
  }
}
signed main() {
  fast_io();
  cin >> a;
  bfs();
}
