Codefroces 595 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.

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.

分为k快，总共为n，所以每块的位数为n/k，且每块只能填a[i]的倍数，且每块不能以b[i]开头，问方案数多少，位数不足0来补

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int n,k,a[100006],b[100006];
ll val[11],ans,vis,pos,cnt;
init()
{
    val[0]=1;
    for(int i=1;i<=10;i++)
        val[i]=val[i-1]*10;
}
int main()
{
    init();
    scanf("%d%d",&n,&k);
    ans=1;
    for(int i=1;i<=n/k;i++)
    {
        scanf("%d",&a[i]);
    }
    for(int i=1;i<=n/k;i++)
    {
        scanf("%d",&b[i]);
    }
    for(int i=1;i<=n/k;i++)
    {
        vis=(val[k]-1)/a[i]+1;
        pos=(val[k-1]*(b[i]+1)-1)/a[i]+1;
        if(b[i]==0) ans=(ans*(vis-pos))%MOD;
        else
        {
            cnt=(val[k-1]*b[i]-1)/a[i]+1;
            ans=(ans*(vis-pos+cnt))%MOD;
        }
    }
    printf("%I64d\n",ans);
    return 0;
}