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;
int d, s;
bool vis[505][5005];
int pre[505][5005][3];
queue<int> qx, qy;
void bfs() {
  vis[0][0] = 1;
  qx.push(0);
  qy.push(0);
  while (!qx.empty()) {
    int x = qx.front();
    qx.pop();
    int y = qy.front();
    qy.pop();
    for (int i = 0; i <= 9; i++) {
      int xx = (x * 10 + i) % d, yy = y + i;
      if (vis[xx][yy]) continue;
      if (yy > s) continue;
      vis[xx][yy] = 1;
      pre[xx][yy][0] = x;
      pre[xx][yy][1] = y;
      pre[xx][yy][2] = i;
      qx.push(xx);
      qy.push(yy);
    }
  }
}
void out(int a, int b) {
  if (a == 0 && b == 0) return;
  out(pre[a][b][0], pre[a][b][1]);
  cout << pre[a][b][2];
}
int main() {
  cin >> d >> s;
  bfs();
  if (vis[0][s] == 0)
    printf("-1");
  else
    out(0, s);
  return 0;
}