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 <cstdint>
using namespace std;

const int64_t MOD = 1000000007;

int N, L;
int64_t x[5050], a[5050];

int main() {
    scanf("%d%d", &N, &L);
    for (int j = 0; j < N; ++j) { scanf("%lld", x+j); }
    for (int j = 0; j < N; ++j) { scanf("%lld", a+j); }

    int64_t memo[5050];

    fill(memo, memo+L, 0);
    if (a[N-1] == 0) { memo[x[N-1]] = 1; }
    else { for (int j = x[N-1]; j < L; j += a[N-1]) { memo[j] = 1; } }

    for (int j = N-2; j >= 0; --j) {
        int64_t acc = 0;
        for (int k = L-1; k >= x[j]; --k) {
            int temp = memo[k];
            if ( (a[j] == 0 && k == x[j]) || (a[j] != 0 && (k - x[j]) % a[j] == 0) ) {
                memo[k] = acc % MOD;
            } else {
                memo[k] = 0;
            }

            acc = (acc + temp) % MOD;
        }
    }

    int64_t ans = 0;
    for (int j = 0; j < L; ++j) {
        ans = (ans + memo[j]) % MOD;
    }
    cout << ans << endl;


    return 0;
}