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;
const int sz = 5e2 + 10, sz2 = 5e3 + 10, inf = 1e9;
int dp[sz][sz2];
bool us[sz][sz2];
vector<pair<int, int> > sv[sz][sz2];
void dfs(int d, int s) {
  us[d][s] = 1;
  for (int a = 0; a < sv[d][s].size(); a++) {
    int nd = sv[d][s][a].first, ns = sv[d][s][a].second;
    if (us[nd][ns] == 0) dfs(nd, ns);
  }
}
int main() {
  int d, s;
  cin >> d >> s;
  for (int a = 0; a < d; a++)
    for (int b = 0; b <= s; b++) dp[a][b] = inf;
  dp[0][0] = inf;
  queue<pair<int, int> > qu;
  qu.push({0, 0});
  while (qu.size()) {
    int cd = qu.front().first, cs = qu.front().second;
    for (int i = 0; (i < 10 and cs + i <= s); i++) {
      int nd = (cd * 10 + i) % d, ns = cs + i;
      if (dp[nd][ns] == inf) dp[nd][ns] = dp[cd][cs] + 1, qu.push({nd, ns});
      if (dp[nd][ns] == dp[cd][cs] + 1) sv[nd][ns].push_back({cd, cs});
    }
    qu.pop();
  }
  if (dp[0][s] != inf) {
    dfs(0, s);
    int cd = 0, cs = 0;
    while (!(cd == 0 and cs == s)) {
      for (int i = 0; i < 10; i++) {
        int nd = (cd * 10 + i) % d, ns = cs + i;
        if (us[nd][ns] and dp[nd][ns] == dp[cd][cs] + 1) {
          printf("%d", i);
          cd = nd, cs = ns;
          break;
        }
      }
    }
  } else
    cout << -1;
}
