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<vector>
using namespace std;
const int mod = 1e9 + 7;

int main(){
    int n, l;
    cin >> n >> l;
    vector<int> x(n), a(n);
    for(int i = 0; i < n; i++)  cin >> x[n-1-i];
    for(int i = 0; i < n; i++)  cin >> a[n-1-i];
    vector<vector<int>> dp(2, vector<int>(l+1, 0));
    for(int j = x[0]; j < l; j += a[0]){
        dp[0][j] = 1;
        if(a[0] == 0)   break;
    }
    for(int j = l-1; j >= 0; j--)  dp[0][j] = (dp[0][j] + dp[0][j+1])%mod;
    for(int i = 1; i < n; i++){
        fill(dp[i%2].begin(), dp[i%2].end(), 0);
        for(int j = x[i]; j < l; j += a[i]){
            dp[i%2][j] = dp[(i+1)%2][j+1];
            if(a[i] == 0)   break;
        } 
        for(int j = l-1; j >= 0; j--)   dp[i%2][j] = (dp[i%2][j] + dp[i%2][j+1])%mod;
    }
    cout << dp[(n+1)%2][0] << endl;
    return 0;
}