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;
int dp[6100][610];
int mul[6100 * 610];
bool used[6100][610];
int len;
int xx[6100 * 610], yy[6100 * 610];
void out(int x, int y) {
  if (dp[x][y] == 0) return;
  bool Find = false;
  for (int i = 0; i < 10 && i <= x && !Find; i++) {
    if (dp[x - i][(((y - mul[dp[x][y] - 1] * i) % d) + d) % d] <=
        dp[x][y] - 1) {
      printf("%d", i);
      int diff =
          dp[x][y] - dp[x - i][(((y - mul[dp[x][y] - 1] * i) % d) + d) % d];
      for (int j = 1; j < diff; j++) printf("0");
      Find = true;
      out(x - i, ((y - mul[dp[x][y] - 1] * i) % d + d) % d);
    }
  }
  assert(Find);
}
int main() {
  scanf("%d%d", &d, &s);
  mul[0] = 1;
  for (int i = 1; i <= d * s; i++) mul[i] = 10 * mul[i - 1] % d;
  dp[0][0] = 0;
  int q = 1, h = 1;
  xx[1] = 0;
  yy[1] = 0;
  used[0][0] = true;
  while (q <= h) {
    for (int i = 0; i < 10; i++) {
      int X = xx[q] + i, Y = (yy[q] * 10 + i) % d;
      if (X <= s && !used[X][Y]) {
        used[X][Y] = true;
        dp[X][Y] = dp[xx[q]][yy[q]] + 1;
        h++;
        xx[h] = X;
        yy[h] = Y;
      }
    }
    q++;
  }
  if (!used[s][0]) {
    printf("-1\n");
  } else {
    out(s, 0);
  }
}
