codeforce 1478B B. Nezzar and Lucky Number 找规律 打表 C

codeforce 1478B B. Nezzar and Lucky Number 找规律 打表 C

https://codeforces.com/contest/1478/problem/B

B. Nezzar and Lucky Number

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

Nezzar's favorite digit among 1,…,91,…,9 is dd. He calls a positive integer lucky if dd occurs at least once in its decimal representation.

Given qq integers a1,a2,…,aqa1,a2,…,aq, for each 1≤i≤q1≤i≤q Nezzar would like to know if aiai can be equal to a sum of several (one or more) lucky numbers.

Input

The first line contains a single integer tt (1≤t≤91≤t≤9) — the number of test cases.

The first line of each test case contains two integers qq and dd (1≤q≤1041≤q≤104, 1≤d≤91≤d≤9).

The second line of each test case contains qq integers a1,a2,…,aqa1,a2,…,aq (1≤ai≤1091≤ai≤109).

Output

For each integer in each test case, print "YES" in a single line if aiai can be equal to a sum of lucky numbers. Otherwise, print "NO".

You can print letters in any case (upper or lower).

Example

input

Copy

2
3 7
24 25 27
10 7
51 52 53 54 55 56 57 58 59 60

output

Copy

YES
NO
YES
YES
YES
NO
YES
YES
YES
YES
YES
YES
NO

Note

In the first test case, 24=17+724=17+7, 2727 itself is a lucky number, 2525 cannot be equal to a sum of lucky numbers.

分析

题目需要判断一个数能否由几个幸运数组成，但a数组的范围是1e9

猜测大数据直接被屏蔽，可以得到d*10之后就都可以任意表示，因为10可以任意使用，个位也可以任意使用

小数据直接疯狂暴力即可

比赛的时候因为数组范围又疯狂wa了两次

代码

https://codeforces.com/contest/1478/problem/B

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <sstream>
#include <iostream>
#include <time.h>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <string.h>
#include <bitset>
#define sf scanf
#define pf printf
#define lf double
#define p123 printf("123
");
#define pn printf("
");
#define pk printf(" ");
#define p(n) printf("%d",n);
#define pln(n) printf("%d
",n);
#define s(n) scanf("%d",&n);
#define ss(n) scanf("%s",n);
#define ps(n) printf("%s",n);
#define sld(n) scanf("%lld",&n);
#define pld(n) printf("%lld",n);
#define slf(n) scanf("%lf",&n);
#define plf(n) printf("%lf",n);
#define sc(n) scanf("%c",&n);
#define pc(n) printf("%c",n);
#define gc getchar();
#define ll long long
#define re(n,a) memset(n,a,sizeof(n));
#define len(a) strlen(a)
#define eps 1e-13
#define zero(x) (((x) > 0? (x):(-x)) < eps)
using namespace std;

int temp[15][20000];
int main(){
    int t;
    s(t)
    re(temp,0);
    for(int i = 1; i<= 10000; i ++){
        temp[1][i] = 1;
    }
    for(int i = 2; i <= 9; i ++){
        for(int j = 1; j <= 30; j ++){
            for(int k = 0; k <= 30; k ++){
                temp[i][i*j+k*10] = 1;
            }
        }
    }
    while(t --){
        int n,d;
        s(n) s(d)
        int x;
        for(int i = 0; i < n; i ++){
            s(x)
            if(x >= d*10){
                puts("YES");
            }else if(temp[d][x] == 1){
                puts("YES");
            }else{
                puts("NO");
            }

        }


    }
    return 0;
}