A. Find a Number

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two positive integers dd and ss. Find minimal positive integer nn which is divisible by dd and has sum of digits equal to ss.

Input

The first line contains two positive integers dd and ss (1≤d≤500,1≤s≤50001≤d≤500,1≤s≤5000) separated by space.

Output

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

Examples

input

Copy

13 50

output

Copy

699998

input

Copy

61 2

output

Copy

1000000000000000000000000000001

input

Copy

15 50

output

Copy

-1

題意：找到一個數，使得能被d整除，且每一位的和是s

解題思路：記憶化寬搜。用二維陣列記錄vis[i][j]代表當前餘數為i，和為j是否訪問過。因為是寬搜，所以找到的一定是最小的。

為了得到最小，要從0~9遍歷。然後用一個數組記錄路徑即可。

#include <bits/stdc++.h>
using namespace std;
const int M=5005;
const int N=505;
const int INF=0x3f3f3f3f;
int d,s;
bool vis[N][M];
int pre[N][M][3];//用於輸出路徑

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=(10*x+i)%d,yy=y+i;
            if(vis[xx][yy]||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 x,int y) {
    if(x==0&&y==0)
        return;
    out(pre[x][y][0],pre[x][y][1]);
    printf("%d",pre[x][y][2]);
}

int main() {
    cin>>d>>s;
    bfs();
    if(vis[0][s]==0)
        puts("-1");
    else
        out(0,s);
    return 0;
}