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:
// Template {{{
#include <bits/stdc++.h>
#define REP(i,n) for(int i=0; i<(int)(n); ++i)
using namespace std;
typedef long long LL;

template<typename T1, typename T2>
ostream& operator << (ostream& os, const pair<T1, T2>& p) {
    os << "(" << p.first << ", " << p.second << ")";
    return os;
}

template<typename T>
ostream& operator << (ostream& os, const vector<T>& v) {
    os << "{";
    for(int i = 0; i < v.size(); i++)
        os << v[i] << (i+1<v.size()?", ":"");
    os << "}";
    return os;
}

class range {
  struct Iterator {
    int val, inc;
    int operator*() {return val;}
    bool operator!=(Iterator& rhs) {return val < rhs.val;}
    void operator++() {val += inc;}
  };
  Iterator i, n;
public:
  range(int e) : i({0, 1}), n({e, 1}) {}
  range(int b, int e) : i({b, 1}), n({e, 1}) {}
  range(int b, int e, int inc) : i({b, inc}), n({e, inc}) {}
  Iterator& begin() {return i;}
  Iterator& end() {return n;}
};

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
inline bool valid(int x, int w) { return 0 <= x && x < w; }

void iostream_init() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.setf(ios::fixed);
    cout.precision(12);
}
//}}}

static const unsigned MODVAL = 1000000007;
struct mint {
    unsigned val;
    mint():val(0){}
    mint(int      x):val(x%MODVAL) {}
    mint(unsigned x):val(x%MODVAL) {}
    mint(LL       x):val(x%MODVAL) {}
};
mint& operator+=(mint& x, mint y) { return x = x.val+y.val; }
mint& operator-=(mint& x, mint y) { return x = x.val-y.val+MODVAL; }
mint& operator*=(mint& x, mint y) { return x = LL(x.val)*y.val; }
mint operator+(mint x, mint y) { return x+=y; }
mint operator-(mint x, mint y) { return x-=y; }
mint operator*(mint x, mint y) { return x*=y; }
int main(){
    iostream_init();
    int N, L;
    while(cin >> N >> L) {
        vector<int> x(N);
        vector<int> a(N);
        REP(i, N) cin >> x[i];
        REP(i, N) cin >> a[i];
        vector<mint> dp(L);
        vector<mint> sum(L+1);
        for(int d = x[0]; d < L; d += a[0]) {
            dp[d] = 1;
            if(a[0] == 0) break;
        }
        REP(i, L){
            sum[i+1] = sum[i] + dp[i];
        }

        for(int k = 1; k < N; k++) {
            vector<mint> ndp(L);

            for(int d = x[k]; d < L; d += a[k]) {
                ndp[d] += sum[d];
                if(a[k] == 0) break;
            }

            dp.swap(ndp);
            sum[0] = 0;
            REP(i, L){
                sum[i+1] = sum[i] + dp[i];
            }
        }

        cout << sum[L].val << endl;
    }

    return 0;
}

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