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:
#define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_DEPRECATE

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <cfloat>
#include <ctime>
#include <cassert>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <numeric>
#include <list>
#include <iomanip>
#include <fstream>
#include <iterator>
#include <bitset>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> Pii;
typedef pair<ll, ll> Pll;

#define FOR(i,n) for(int i = 0; i < (n); i++)
#define sz(c) ((int)(c).size())
#define ten(x) ((int)1e##x)
#define tenll(x) ((ll)1e##x)

int x[5000], a[5000];
int dp[2][5010];
const int MOD = ten(9) + 7;

int main(){
	int n, l;
	scanf("%d%d", &n, &l);
	FOR(i, n) scanf("%d", x + i);
	FOR(i, n) scanf("%d", a + i);
	FOR(i, n) x[i]++;

	FOR(i,l+1) dp[0][i] = 1;
	auto pv = dp[0], nt = dp[1];
	FOR(i, n){
		memset(nt, 0, sizeof(dp[0]));
		if (a[i]) {
			for (int dist = 0; x[i] + dist <= l; dist += a[i]) {
				int cur = x[i] + dist;
				nt[cur] = pv[cur - 1];
			}
		} else {
			nt[x[i]] = pv[x[i] - 1];
		}
		FOR(j, l){
			nt[j + 1] += nt[j];
			if (nt[j + 1] >= MOD)  nt[j + 1] -= MOD;
		}
		swap(pv, nt);
	}
	cout << pv[l] << endl;
}