Codeforces 595 B. Pasha and Phone (容斥）

题目连接：http://codeforces.com/contest/595/problem/B

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, ..., an / k and b1, 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 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 bi and
is divisible by ai 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,...,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 length n,
for the given k, ai and bi.
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, ..., an / k (1 ≤ ai < 10k).

The third line of the input contains n / k space-separated positive integers — sequence b1, b2, ..., bn / k (0 ≤ bi ≤ 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.

题目大意：

给你ai和bi，让你找到有多少个k位数，使得这个k位数不以bi开头且mod ai=0
处理n/k次，然后把所有的答案都乘起来。

解题思路：mid ai=0 的方案数减去bi开头的方案数。对bi=0的情况分类讨论。

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┆　　┃　勒　┃　　┆　　　　　　
┆　　┃　戈　┗━━━┓ ┆
┆　　┃　壁　　　　　┣┓┆
┆　　┃　的草泥马　　┏┛┆
┆　　┗┓┓┏━┳┓┏┛ ┆
┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <bitset>
using namespace std;

#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)
#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)
#define pb push_back
#define mp make_pair
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define LL long long
#define ULL unsigned long long
#define MS0(X) memset((X), 0, sizeof((X)))
#define SelfType int
SelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}
SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}
#define Sd(X) int (X); scanf("%d", &X)
#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())
#define all(a) a.begin(), a.end()
#define mem(x,v) memset(x,v,sizeof(x))
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<int> vi;
typedef vector<long long> vll;
inline LL read(){LL ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")

LL a[100005];
LL b[100005];
LL f[10];

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
ios::sync_with_stdio(0);
cin.tie(0);
int n,k;
n = read(),k = read();
f[0] = 1;
for(int i=1;i<=n/k;i++) a[i] = read();
for(int i=1;i<=n/k;i++) b[i] = read();
for(int i=1;i<=10;i++) f[i] = f[i-1]*10;
LL ans = 1;
for(int i=1;i<=n/k;i++)
{
LL t1 = (f[k]-1)/a[i] + 1; //是a[i]的倍数的数的个数。+1是全是0的情况
LL t2 = (f[k-1]-1)/a[i] + 1; //是a[i]的0倍的数的个数。
LL t3 = (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 *= t1 - t2;
else ans *= t1 - t3;
ans %= mod;
}
printf("%I64d\n",ans);
return 0;
}