Codeforces Round #330 (Div. 2) B. Pasha and Phone

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, ..., 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.

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：长度为n的串 每长度k为一个block（块）

 第二行输入 n/k个数 表示  第i块中的数必须是a[i]的倍数

 第三行输入 n/k个数 表示 第i块的首位不能为b[i]

注意 首位为0的处理！！！

还是水果A题 手速不行 B 字符串题目 要多练了

#include<bits/stdc++.h>
using namespace std;
#define N 1000000007    
__int64 n,k;
__int64 a[100000];
__int64 b[100000];
__int64 re;
int main()
{
    scanf("%I64d%I64d",&n,&k);
    re=1;
    for(__int64 i=1;i<=n/k;i++)
        scanf("%I64d",&a[i]);
    for(__int64 i=1;i<=n/k;i++)
        scanf("%I64d",&b[i]);
    for(__int64 i=1;i<=n/k;i++)
    {
       __int64 exm=1;
       __int64 linshi=0;
       for(int j=1;j<=k;j++)
           exm*=10;
       if(b[i]!=0)
       linshi=linshi+(exm-1)/a[i]+1-(((b[i]+1)*(exm/10)-1)/a[i]-((b[i])*(exm/10)-1)/a[i]);
        else
       linshi=linshi+(exm-1)/a[i]-(exm/10-1)/a[i];
       re=re*linshi;
       re=re%N;
    }
    printf("%I64d",re);
    return 0;
}