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<bits/stdc++.h>
using namespace std;
typedef long long ll;

ll dp[5005][5005];
ll N,L;
ll x[5005];
ll a[5005];

const ll mod = 1000000007LL;

void add( ll &a, ll b ){
  a += b;
  a %= mod;
}

ll solve(int id,int l){
  //cout << id << ", "<< l << " "<< x[id] << " "<<a[id] << endl;
  if( dp[id][l] != -1 ) return dp[id][l];
  ll ret = 0;  
  if( id == 0 ) return dp[id][l] = 1LL;
  if( x[id] > l ) return dp[id][l] = 0;
                    
  if( x[id] == l || (a[id] > 0 && (l - x[id])%a[id] == 0) ) add( ret, solve(id-1,l-1) );
  add( ret, solve( id, l-1 ) );
  return dp[id][l] = ret;
}

int main(){
  memset(dp,-1,sizeof(dp));
  cin >> N >> L;
  for(int i=0;i<N;i++){
    cin >> x[i+1]; x[i+1]++;
  }
  for(int i=0;i<N;i++)
    cin >> a[i+1]; 
  cout << solve( N, L ) << endl;
}