CodeForces - 595B Pasha and Phone （数学技巧）好题

64bit IO Format: %I64d & %I64u

CodeForces - 595B

Pasha and Phone

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Description

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

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_iif 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 Input

Input

6 238 56 497 3 4

Output

8

Input

8 21 22 3 445 4 3 2

Output

32400

Hint

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

//题意：给你一个n,k,表示要得到一个长度为n的号码，接下来给长度为n/k的a[],b[].

要得到一个n位数的号码，并且这个号码分为n/k块，在第i块的数字，只能是a[i]的倍数，并且不能以b[i]开头，问这样的号码有几个。

思路：

直接模拟，（参考大神的^_^）..

#include<stdio.h>  
#include<string.h>  
#include<math.h>  
#include<algorithm>  
#include<iostream>  
#include<queue>  
#define INF 0x3f3f3f3f  
#define IN __int64  
#define ull unsigned long long  
#define ll long long  
#define N 100010  
#define M 1000000007  
using namespace std;
ll f[10];
void getfac()
{
	f[0]=1;
	for(int i=1;i<10;i++)
		f[i]=f[i-1]*10;
}
ll a[N],b[N];
int main()
{
	getfac();
	int n,m,k,i,j;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		for(i=0;i<n/k;i++)
			scanf("%d",&a[i]);
		for(i=0;i<n/k;i++)
			scanf("%d",&b[i]);
		ll ans=1;
		for(i=0;i<n/k;i++)
		{
			ll num1=(f[k]-1)/a[i]+1;//是a[i]的倍数的数的个数。 
			ll num2=(f[k-1]-1)/a[i]+1;//是a[i]的0倍的数的个数 
			ll num3=(f[k-1]*(b[i]+1)-1)/a[i]-(f[k-1]*b[i]-1)/a[i];//是a[i]的倍数并且包含b[i]的数的个数 
			if(b[i]==0)
				ans*=num1-num2;
			else
				ans*=num1-num3;
			ans%=M;
		}
		printf("%lld\n",ans);
	}
	return 0;
}