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
#include <bits/stdc++.h>
using namespace std;
int d, s, Pr[5010][555][2];
struct node {
  int a, b;
  node() {}
  node(int _1, int _2) {
    a = _1;
    b = _2;
  }
};
int main() {
  scanf("%d%d", &d, &s);
  vector<node> V;
  V.clear();
  V.push_back(node(0, 0));
  memset(Pr, -1, sizeof Pr);
  Pr[0][0][0] = 0;
  for (int i = 0; Pr[s][0][0] == -1 && i < V.size(); i++) {
    int a = V[i].a, b = V[i].b;
    if (a > s) continue;
    for (int j = 0; j < 10; j++)
      if (Pr[a + j][(b * 10 + j) % d][0] == -1) {
        Pr[a + j][(b * 10 + j) % d][0] = j;
        Pr[a + j][(b * 10 + j) % d][1] = b;
        V.push_back(node(a + j, (b * 10 + j) % d));
      }
  }
  if (Pr[s][0][0] == -1)
    puts("-1");
  else {
    vector<int> V;
    int a = s, b = 0;
    while (a != 0) {
      int tmpb = Pr[a][b][1];
      V.push_back(Pr[a][b][0]);
      a -= Pr[a][b][0];
      b = tmpb;
    }
    for (int i = (int)V.size() - 1; i >= 0; i--) printf("%d", V[i]);
  }
  return 0;
}
