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;

#define MAX 5050
typedef long long ll;
const ll MOD = ((1e9) + 7);

int dp[2][MAX];

int main()
{
    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];
    }

    for (int i = 0; i < L; i++) {
        dp[0][i] = 1;
    }
    
    for (int i = 0; i < N; i++) {
        int curr = i & 1, next = !curr;
        for (int j = x[i]; j < L; j += a[i]) {
            if (j > 0) {
                dp[next][j] += dp[curr][j-1];
            } else {
                dp[next][j] = dp[curr][j];
            }
            dp[next][j] %= MOD;
            if (a[i] == 0) break;
        }
        
        for (int j = x[i]; j < L; j++) {
            dp[next][j+1] += dp[next][j];
            dp[next][j+1] %= MOD;
        }
        fill(dp[curr], dp[curr] + L, 0);
    }
    
    cout << dp[N&1][L-1] << endl;
    return 0;
}