Problem: You are given two positive integers d and s. Find minimal positive integer n which is divisible by d and has sum of digits equal to s.

Input

The first line contains two positive integers d and s (1 ≤ d ≤ 500, 1 ≤ s ≤ 5000) separated by space.

Output

Print the required number or -1 if it doesn't exist.

Examples

Input

13 50


Output

699998


Input

61 2


Output

1000000000000000000000000000001


Input

15 50


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 5000 + 5;
const int MOD = 500 + 5;
int dp[N][MOD];
pair<int, int> ant[N][MOD];
queue<pair<int, int>> q;
int main() {
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < MOD; j++) {
      dp[i][j] = -1;
    }
  }
  int mod, tot;
  cin >> mod >> tot;
  for (int i = 1; i <= 9; i++) {
    q.push({i, i % mod});
    dp[i][i % mod] = i;
  }
  while (!q.empty()) {
    int s0 = q.front().first, r0 = q.front().second;
    q.pop();
    for (int i = 0; i <= 9; i++) {
      int s = s0 + i;
      int r = (10 * r0 + i) % mod;
      if (s > tot || dp[s][r] != -1) continue;
      dp[s][r] = i;
      ant[s][r] = {s0, r0};
      q.push({s, r});
    }
  }
  if (dp[tot][0] == -1) {
    cout << "-1\n";
    return 0;
  }
  vector<int> ans;
  int s = tot;
  int r = 0;
  while (s) {
    ans.push_back(dp[s][r]);
    pair<int, int> rof = ant[s][r];
    s = rof.first;
    r = rof.second;
  }
  reverse(ans.begin(), ans.end());
  for (auto &x : ans) {
    cout << x;
  }
  cout << "\n";
  return 0;
}