Codeforces 595B. Pasha and Phone 容斥

B. 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 n digits.

Also Pasha has a number k and two sequences of length n / k (n is divisible by k) a1, a2, ..., an / k and b1, b2, ..., bn / 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 bi and is divisible by ai 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,...,ck. Then the integer is calculated as the result of the expression c1·10k - 1 + c2·10k - 2 + ... + ck.

Pasha asks you to calculate the number of good phone numbers of length n, for the given k, ai and bi. 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, ..., an / k (1 ≤ ai < 10k).

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

Output

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

Sample test(s)

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.

 

题意：输入n，k接下来2行输入a₁，a₂,...a_n/k和b₁，b₂,...b_{n/k。电话号码由n/k段组成，每段有k个数字。每段电话号码的数字要为a[i]的倍数，且不能以b[i]开头。如果不够k为就前补0，那么就是0开头，如果b不为0的话，那段数字可以全为0。输出号码有几种可能性。}

思路：利用容斥，每段号码的不考虑b的情况下可能性有gg=(pow(10,k)-1)/a[i]种，在减去开头是b的情况。

 

#include<bits/stdc++.h>
using namespace std;
int a[100000],b;
int main()
{
    int i,n,k;
    scanf("%d%d",&n,&k);
    for(i=0; i<n/k; i++)
        scanf("%d",&a[i]);
    __int64 gg,ans=1;
    int sign=k,flag=1;
    while(sign--) flag*=10;
    for(i=0; i<n/k; i++)
    {
        scanf("%d",&b);
        gg=(flag-1)/a[i];
        if(b!=0)
        {
            gg++;
            gg-=((b+1)*(flag/10)-1)/a[i]-(b*(flag/10)-1)/a[i];
        }
        else gg-=(flag/10-1)/a[i];
        ans=(ans*gg)%1000000007;
    }
    cout<<ans<<endl;
    return 0;
}