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 <string>
#include <vector>
#include <algorithm>
#include <cstring>

using namespace std;

static const int MOD = 1000000007;
static int dp[5001][5001];

int main(){
	ios_base::sync_with_stdio(false);
	int n, l;
	cin >> n >> l;
	vector<int> x(n), a(n);
	for(int i = 0; i < n; ++i){ cin >> x[i]; }
	for(int i = 0; i < n; ++i){ cin >> a[i]; }
	memset(dp, 0, sizeof(dp));
	dp[n][l] = 1;
	for(int i = n - 1; i >= 0; --i){
		int sum = dp[i + 1][l];
		for(int j = l - 1; j >= x[i]; --j){
			if(a[i] == 0){
				if(j == x[i]){ dp[i][j] = (dp[i][j] + sum) % MOD; }
			}else if((j - x[i]) % a[i] == 0){
				dp[i][j] = (dp[i][j] + sum) % MOD;
			}
			sum = (sum + dp[i + 1][j]) % MOD;
		}
	}
	int answer = 0;
	for(int i = 0; i <= l; ++i){
		answer = (answer + dp[0][i]) % MOD;
	}
	cout << answer << endl;
	return 0;
}