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 <bits/stdc++.h>
using namespace std;
using ll = long long;
#define int ll
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define all(c) begin(c), end(c)
#define dump(x) cerr << __LINE__ << ":\t" #x " = " << (x) << endl

const int MOD = 1000000007;
int n, l;
int x[5010], a[5010];

int dp[5010][5010];

signed main(){
    while(cin >> n >> l){
        rep(i, n) cin >> x[i], ++x[i];
        rep(i, n) cin >> a[i], a[i] ? 0 : a[i] = l;
        memset(dp, 0, sizeof dp);
        for(int i = 0; i <= l; ++i) dp[0][i] = 1;
        for(int i = 1; i <= n; ++i){
            int a = ::a[i-1], x = ::x[i-1];
            for(int j = x; j <= l; ++j){
                if(j == x){
                    dp[i][j] = dp[i-1][j-1];
                } else {
                    if((j-x)%a == 0){
                        dp[i][j] += dp[i-1][j-1];
                    }
                    dp[i][j] += dp[i][j-1];
                    dp[i][j] %= MOD;
                }
            }
        }

        // for(int i = 0; i <= n; ++i){
        //     for(int j = 0; j <= l; ++j){
        //         cout << dp[i][j] << ' ';
        //     }
        //     cout << endl;
        // }

        cout << dp[n][l] << endl;
    }
}