Pasha and Phone(思维)

1. Pasha and Phone

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Pasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consists of exactly ndigits.

Also Pasha has a number k and two sequences of length n / k (n is divisible by k) a₁, a₂, ..., a_n / k and b₁, b₂, ..., b_n / k. Let's split the phone number into blocks of length k. The first block will be formed by digits from the phone number that are on positions 1, 2,..., k, the second block will be formed by digits from the phone number that are on positions k + 1, k + 2, ..., 2·k and so on. Pasha considers a phone number good, if the i-th block doesn't start from the digit b_i and is divisible by a_i if represented as an integer.

To represent the block of length k as an integer, let's write it out as a sequence c₁, c₂,...,c_k. Then the integer is calculated as the result of the expression c₁·10^k - 1 + c₂·10^k - 2 + ... + c_k.

Pasha asks you to calculate the number of good phone numbers of length n, for the given k, a_i and b_i. As this number can be too big, print it modulo 10⁹ + 7.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ min(n, 9)) — the length of all phone numbers and the length of each block, respectively. It is guaranteed that n is divisible by k.

The second line of the input contains n / k space-separated positive integers — sequence a₁, a₂, ..., a_n / k (1 ≤ a_i < 10^k).

The third line of the input contains n / k space-separated positive integers — sequence b₁, b₂, ..., b_n / k (0 ≤ b_i ≤ 9).

Output

Print a single integer — the number of good phone numbers of length n modulo 10⁹ + 7.

Examples

input

6 2 38 56 49 7 3 4

output

8

input

8 2 1 22 3 44 5 4 3 2

output

32400

Note

In the first test sample good phone numbers are: 000000, 000098, 005600, 005698, 380000, 380098, 385600, 385698.

题解：小明要记电话号码了，给两个整数n,k，接下来是两个数组，有n/k个元素，让找是a数组对应块的倍数，又不能是b[i]开头的电话号的种类数；

很容易想到分成k块，分别找对应块的种类数；想着就去做了，假设k=2；b[i]=5,当对应块是5的时候也是符合的；因为是05，0开头的，错了几次，好像cf里面没有log

函数还是别的原因，我用log10（x）取小数找首数字的方法老是出问题，自己编译器上就没事，但是最后发现如果能用也会超时；暴力肯定解决不了；

因为是倍数；每个块里面可能有pow(10,k)个数，直接求出是a[i]倍数数字的个数，再减去b[i]**开头里面所占a[i]倍数的个数就是答案了；由于b[i]可能为0；所以

还要单独判断b[i]=0的情况，b[i]**开头里面所占a[i]倍数的个数就是b[i]+1开头所占a[i]倍数的个数减去b[i]开头所占a[i]的个数；

有点麻烦，需要考虑全面；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<string>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define mem(x,y) memset(x,y,sizeof(x))
typedef __int64 LL;
const int MAXN=;
const int MOD=1e9+;
int a[MAXN],b[MAXN],c[MAXN];
int main(){
    int n,k;
    while(~scanf("%d%d",&n,&k)){
        mem(c,);
        int p[];p[]=;
        for(int i=;i<;i++)p[i]=p[i-]*;
        for(int i=;i<n/k;i++)SI(a[i]);
        for(int j=;j<n/k;j++)SI(b[j]);
        for(int i=;i<n/k;i++){
            if(a[i]==){
                if(b[i]!=)c[i]++;continue;
            }
            /*
            else for(int j=0;;j++){
                if(j*a[i]>=p[k])break;
                //int t=pow(10,log10(1.0*j*a[i])-(int)log10(1.0*j*a[i]));
                if(j*a[i]/p[k-1]!=b[i])c[i]++;
            }
            */
            int x1,x2,x3;
            x1=(p[k]-)/a[i]+;
            x2=(p[k-]-)/a[i]+;
            x3=(p[k-]*(b[i]+)-)/a[i]-(p[k-]*b[i]-)/a[i];
            if(b[i]==)c[i]=x1-x2;
            else c[i]=x1-x3;
        }
        LL ans=;
        for(int i=;i<n/k;i++){
            ans=ans*c[i]%MOD;
        }
        printf("%I64d\n",ans);
    }
    return ;
}