Problem: Problem statement

N-winged rabbit is on a balance beam of length L-1. The initial position of the i-th rabbit is the integer x_i, which satisfies 0 ≤ x_ {i} \ lt x_ {i + 1} ≤ L−1. The coordinates increase as you move to the right. Any i-th rabbit can jump to the right (ie, move from x_i to x_i + a_i) any number of times, just a distance a_i. However, you cannot jump over another rabbit or enter a position below -1 or above L. Also, at most one rabbit can jump at the same time, and at most one rabbit can exist at a certain coordinate.

How many possible states of x_ {0},…, x_ {N−1} after starting from the initial state and repeating the jump any number of times? Find by the remainder divided by 1 \, 000 \, 000 \, 007.

input

The input is given in the following format.


N L
x_ {0}… x_ {N−1}
a_ {0}… a_ {N−1}


Constraint

* All inputs are integers
* 1 \ ≤ N \ ≤ 5 \,000
* N \ ≤ L \ ≤ 5 \,000
* 0 \ ≤ x_ {i} \ lt x_ {i + 1} \ ≤ L−1
* 0 \ ≤ a_ {i} \ ≤ L−1



output

Print the answer in one line.

sample

Sample input 1


13
0
1


Sample output 1


3


If 1/0 is used to express the presence / absence of a rabbit, there are three ways: 100, 010, and 001.

Sample input 2


twenty four
0 1
1 2


Sample output 2


Four


There are four ways: 1100, 1001, 0101, 0011.

Sample input 3


10 50
0 1 2 3 4 5 6 7 8 9
1 1 1 1 1 1 1 1 1 1


Sample output 3


272278100


The binomial coefficient C (50,10) = 10 \, 272 \, 278 \, 170, and the remainder obtained by dividing it by 1 \, 000 \, 000 \, 007 is 272 \, 278 \, 100.





Example

Input

1 3
0
1


Output

3

Solution (in CPP):

Solution:
#include<stdio.h>
#include<algorithm>
using namespace std;
long long mod=1000000007;
int c[5100];
int d[5100];
int dp[5100][5100];
int sum[5100][5100];
int main(){
	int a,b;
	scanf("%d%d",&a,&b);
	for(int i=0;i<a;i++)scanf("%d",c+i);
	for(int i=0;i<a;i++)scanf("%d",d+i);
	for(int i=c[0];i<b;i+=d[0]){
		dp[1][i]=1;
		if(d[0]==0)break;
	}
	sum[1][0]=dp[1][0];
	for(int i=1;i<b;i++){
		sum[1][i]=(dp[1][i]+sum[1][i-1])%mod;
	}
	for(int i=1;i<a;i++){
		for(int j=c[i];j<b;j+=d[i]){
			if(j)dp[i+1][j]=sum[i][j-1];
			if(d[i]==0)break;
		}
		for(int j=1;j<b;j++)sum[i+1][j]=(dp[i+1][j]+sum[i+1][j-1])%mod;
	}
	printf("%lld\n",sum[a][b-1]);
}