Codeforces Round #330 (Div. 2) B 容斥原理

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

Input

8 2
1 22 3 44
5 4 3 2

Output

Note

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

题意：长度为n的手机号按照每k位分块 n能够除尽k

第i块必须是ai的倍数并且首位不能是bi 问有多少种满足条件的手机号

题解：容斥原理处理统计每块满足条件的个数累乘取模输出答案

每块满足条件的个数=是ai的倍数的个数 -（首位为bi并且是ai的倍数的个数）

注意b==0的处理考虑后k-1位

注意计算首位为bi的并且是ai的倍数的个数的技巧

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<algorithm>
#define ll __int64
#define mod 1000000007
#define PI acos(-1.0)
using namespace std;
ll n,k;
ll a[100005];
ll b[100005];
int main()
{
    scanf("%I64d %I64d",&n,&k);
    for(ll i=1;i<=n/k;i++)
    scanf("%I64d",&a[i]);
    for(ll i=1;i<=n/k;i++)
    scanf("%I64d",&b[i]);
    ll ans=1;
    ll exm=1;
    for(int i=1;i<=k;i++)
        exm*=10;
    for(ll i=1;i<=n/k;i++)
    {
        if(b[i]==0)
        ans=((exm-1)/a[i]-(exm/10-1)/a[i])%mod*ans%mod;//考虑后k-1位
        else
        ans=((exm-1)/a[i]-(((b[i]+1)*exm/10-1)/a[i]-(b[i]*exm/10-1)/a[i])+1)%mod*ans%mod;
    }
     printf("%I64d\n",ans);
    return 0;
}

Codeforces Round #330 (Div. 2) B 容斥原理

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