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<vector>
#define REP(i, N) for(int i=0;i<N;++i)
#define REPA(i, N) for(int i=1;i<N;++i)
#define MOD ((int)1e9+7)
using namespace std;
int main(){
    int N, L;
    cin >> N >> L;
    vector<int> sp(N), ran(N);
    REP(i, N){
        cin >> sp[i];
    }
    REP(i, N){
        cin >> ran[i];
    }
    vector<vector<int> > dp(N, vector<int>(L, 0));
    for(int i=sp[0];i<L;i+=ran[0]){
        dp[0][i] = 1;
        if(ran[0]==0)break;
    }
    REPA(i, N){
        for(int j=sp[i];j<L;j+=ran[i]){
            int sum = 0;
            if(ran[i]==0){
                REP(k, sp[i]){
                    sum = (sum + dp[i-1][k])%MOD;
                }
                dp[i][j] = sum;
                break;
            }else if(j==sp[i]){
                REP(k, sp[i]){
                    sum = (sum + dp[i-1][k])%MOD;
                }
                dp[i][j] = sum;
            }else{
                for(int k=j-ran[i];k<j;++k){
                    sum = (sum + dp[i-1][k])%MOD;
                }
                dp[i][j] = (dp[i][j-ran[i]] + sum)%MOD;
            }
        }
    }
    int ans = 0;
    REP(i, L){
        ans = (ans + dp[N-1][i])%MOD;
    }
    cout << ans << endl;
    return 0;
}