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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<string>
#include<stack>
#include<cstdio>
#include<cmath>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> P;
typedef pair<ll,P> P1;

#define fr first
#define sc second
#define mp make_pair
#define pb push_back
#define rep(i,x) for(ll i=0;i<x;i++)
#define rep1(i,x) for(ll i=1;i<=x;i++)
#define rrep(i,x) for(ll i=x-1;i>=0;i--)
#define rrep1(i,x) for(ll i=x;i>0;i--)
#define sor(v) sort(v.begin(),v.end())
#define rev(s) reverse(s.begin(),s.end())
#define lb(vec,a) lower_bound(vec.begin(),vec.end(),a)
#define ub(vec,a) upper_bound(vec.begin(),vec.end(),a)
#define uniq(vec) vec.erase(unique(vec.begin(),vec.end()),vec.end())
#define min_3(a,b,c) min(a,min(b,c))
#define max_3(a,b,c) max(a,max(b,c))
#define mp1(a,b,c) P1(a,P(b,c))
#define pque(a) priority_queue<a>
#define rpque(a) priority_queue<a,vector<a>,greater<a>>

const int INF = 1000000000;
const int dir_4[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
const int dir_8[8][2]={{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};
const int kaijou[10]={1,1,2,6,24,120,720,5040,40320,362880};

const ll M = 1000000007;

int main(){
	static ll n,l;
	static ll x[5002],a[5002];
	scanf("%lld%lld",&n,&l);
	rep(i,n){
		scanf("%lld",&x[i]);
	}
	rep(i,n){
		scanf("%lld",&a[i]);
	}
	
	static ll dp[2][5002] = {};
	int t = 0,t_ = 1;
	dp[t][l] = 1;
	rrep(i,n){
		rep(j,5002){
			dp[t_][j] = 0;
		}
		rrep(j,l){
			dp[t_][j] = dp[t_][j+1] + dp[t][j+1];
			dp[t_][j] %= M;
		}
		rep(j,5002){
			if(j < x[i] || j > l-1)dp[t_][j] = 0;
			else if(a[i] == 0){
				if(j != x[i])dp[t_][j] = 0;
			}
			else {
				if((j-x[i])%a[i] != 0)dp[t_][j] =0;
			}
		}
		swap(t,t_);
	}
	
	/*rep(i,n){
		rep(j,l){
			printf("%lld ",dp[i][j]);
		}puts("");
	}*/
	
	ll ret = 0;
	rep(i,5002){
		ret += dp[t][i];
	}
	cout << ret%M << endl;
}