Codeforces 595 B. Pasha and Phone【数学】

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 exactlyn digits.

Also Pasha has a number k and two sequences of lengthn / k (n is divisible by k)a1, a2, ..., an / k andb1, 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 positions1,2,...,k, the second block will be formed by digits from the phone number that are on positionsk + 1,k + 2, ...,2·k and so on. Pasha considers a phone numbergood, if thei-th block doesn't start from the digitbi and is divisible byai if represented as an integer.

To represent the block of length k as an integer, let's write it out as a sequencec1,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 lengthn, for the givenk,ai andbi. As this number can be too big, print it modulo109 + 7.

Input

The first line of the input contains two integersn andk (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 thatn is divisible byk.

The second line of the input contains n / k space-separated positive integers — sequencea1, a2, ..., an / k (1 ≤ ai < 10k).

The third line of the input contains n / k space-separated positive integers — sequenceb1, b2, ..., bn / k (0 ≤ bi ≤ 9).

Output

Print a single integer — the number of good phone numbers of lengthn modulo109 + 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.

题目大意：

给你一个电话号长度n，再给你一个数k，表示要将电话号分成n/k部分。每一部分长度为10^k。

而且每一部分必须是ai的倍数，并且不能以bi开头不足10^k长度的部分，前补0。

问你能够找到多少种可行方案。

思路：

1、首先我们知道，假如让我们求100以内3的倍数的个数，那么就是100/3；

2、那么我们设定c【i】表示区间范围（10^k）内有多少个a【i】的倍数，并且设定d【i】表示区间范围内去除的以b【i】开头的数字的个数。那么ans【i】=c【i】-d【i】；最后累乘ans【i】即可。

3、那么两个数组的求法：

c【i】=（10^k-1）/a【i】+1（+1是表示加全是0的这种情况）。

d【i】=（（b【i】+1）*(10^(k-1))-1）/a【i】-（b【i】*(10^(k-1))-1）/a【i】

注意，当b【i】==0的时候要特殊判定。

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;#define ll __int64ll a[100002];ll b[100002];ll c[100002];ll d[100002];ll mod=1e9+7;int main(){    int n,k;    while(~scanf("%d%d",&n,&k))    {        int tmp=n/k;        int len=n/tmp;        ll qujian=1;        for(int i=1;i<=len;i++)        {            qujian*=10;        }        for(int i=0;i<tmp;i++)        {           scanf("%I64d",&a[i]);        }        for(int i=0;i<tmp;i++)        {            c[i]=(qujian-1)/a[i]+1;        }        for(int i=0;i<tmp;i++)        {            scanf("%I64d",&b[i]);            b[i]*=qujian;b[i]/=10;        }        for(int i=0;i<tmp;i++)        {            if(b[i]==0)            {                d[i]=(qujian/10-1)/a[i]+1;            }            else            d[i]=(b[i]+qujian/10-1)/a[i]-(b[i]-1)/a[i];        }        for(int i=0;i<tmp;i++)        {            c[i]-=d[i];        }        ll output=1;        for(int i=0;i<tmp;i++)        {            output*=c[i];            output%=mod;        }        printf("%I64d\n",output%mod);    }}