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;
using Int = long long;
using ll = long long;
const ll MOD=1000000007;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b)a=b;};
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b)a=b;};

signed main(){
  ll n,l;
  cin>>n>>l;
  vector<ll> X(n),A(n);
  for(auto &I:X){cin>>I;}
  for(auto &I:A){cin>>I;}
  vector<ll> dp(l+2,0);
  vector<ll> nx(l+2);
  dp[l]=1;
  for(ll i=n-1;i>=0;i--){
    for(auto &I:nx){I=0;}
    for(ll j=l;j>=0;j--){
      dp[j]+=dp[j+1];
      dp[j]%=MOD;
    }
    if(A[i]==0){
      nx[X[i]]=dp[X[i]+1];
    }
    else{
      for(ll j=X[i];j<l;j+=A[i]){
	nx[j]=dp[j+1];
      }
    }
    dp=nx;
  }
  ll ans=0;
  for(auto &I:dp){ans+=I; ans%=MOD;}
  if(ans<0){ans+=MOD;}
  cout<<ans<<endl;

  
  return 0;
}