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 <vector>
#include <numeric>
#include <algorithm>
#include <math.h>
#include <stdio.h>


using namespace std;
#define LL long long
const LL MOD = (LL)1e9+7;
#define FOR(i,s,e) for(i = s; i < e; i++)
LL n,l,i,j,y,key,r,dp[5001][5001],x[5001],a[5001],la[5002];
int main(){
  cin >> n >> l;
  FOR(i,0,n) cin >> x[i];
  FOR(i,0,n) cin >> a[i];
  i = n-1;
  for(r = l-1;r >= 0;r--){
    if (a[i] == 0){
      if (r == x[i]){
        la[i] = r;
        i--;
      }
    }
    else{
      if ((r-x[i])%a[i] == 0){
        la[i] = r;
        i--;
      }
    }
    if (i < 0)break;
  }
  la[n] = l;

  FOR(i,0,n)FOR(j,x[i],la[i+1])dp[i][j] = 0;
  y = x[0];
  while (y < la[1]){
    dp[0][y] = 1;
    y += a[0];
    if (a[0] == 0)break;
  }
  FOR(i,0,la[1]-1)dp[0][i+1] += dp[0][i];
  FOR(i,1,n){
    if (a[i] == 0){
      FOR(j,x[i],la[i+1]){
        dp[i][j] += dp[i][j-1];
        if (j == x[i]) dp[i][j] += dp[i-1][j-1];
        if (dp[i][j] >= MOD)dp[i][j] %= MOD;
      }
    }
    else{
      key = x[i]%a[i];
      FOR(j,x[i],la[i+1]){
        dp[i][j] += dp[i][j-1];
        if (j%a[i] == key) dp[i][j] += dp[i-1][j-1];
        if (dp[i][j] >= MOD)dp[i][j] %= MOD;
      }
    }
  }
  cout << dp[n-1][l-1] << endl;
  return 0;
}