[VJ][数学]Pasha and Phone

Pasha and Phone 

Description

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.

描述：

有一个整数，整数的位数是n，把整数分为 n/k 块，保证 n/k 是整数。

总共有1—n/k 块，要使第 i 块是a[i]的倍数（包括0倍数），并且第一位不是b[i]。

求这样的整数有多少个。

分析：

刚开始写枚举每一块，使那一块的整数是a[i]的倍数，并且第一位不是b[i]。

有排列组合可知，答案就是每一块的个数相乘。

后来TLE了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
long long ans=1,maxx=1;
long long a[100010];
int b[100010];
int main()
{
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n/k;i++)
        cin>>a[i];
    for(int i=1;i<=n/k;i++)
        cin>>b[i];
    for(int i=1;i<k;i++)
        maxx*=10;
    for(int i=1;i<=n/k;i++)
    {
        long long cnt=0;
        for(int j=0; ;j++)
        {
            if(a[i]*j>=maxx*10)    break;
            else{
                if(a[i]*j/maxx!=b[i])
                    cnt++;
                else continue;
            }
        }
        ans=ans*cnt;
        ans=ans%1000000007;
    }
    cout<<ans<<endl;
    return 0;
}

正确解法：

枚举每一块，每一块的最大数/a[i] 就是除了0倍数之外的倍数的个数。

每一块的最大数/10 /a[i]就是第一位是0倍数的个数

两个都加一代表00成立

再找出第一位是b[i]的块的个数

此时，若b[i]==0   num2！=num3，所以要特殊判断！

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
long long ans=1,maxx=1;
long long a[100010];
int b[100010];
int main()
{
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n/k;i++)
        cin>>a[i];
    for(int i=1;i<=n/k;i++)
        cin>>b[i];
    for(int i=1;i<k;i++)
        maxx*=10;
    for(int i=1;i<=n/k;i++)
    {
        long long num1=(maxx*10-1)/a[i]+1;
        long long num2=(maxx-1)/a[i]+1;
        long long num3=(maxx*(b[i]+1)-1)/a[i]-(maxx*b[i]-1)/a[i];
        if(b[i]==0)    ans*=num1-num2;
        else ans*=num1-num3;
        ans%=1000000007;
    }
    cout<<ans<<endl;
    return 0;
}