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 <iomanip>
#include <sstream>
#include <vector>
#include <string>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iterator>
#include <limits>
#include <numeric>
#include <utility>
#include <cmath>
#include <cassert>
#include <cstdio>

using namespace std; using namespace placeholders;

typedef long long LL;
typedef unsigned long long ULL;
typedef vector< int > VI;
typedef vector< vector< int > > VVI;
typedef vector< string > VS;
typedef stringstream SS;
typedef pair< int, int > PII;
typedef vector< pair< int, int > > VPII;
#define VT( T ) vector< T >
#define VVT( T ) vector< vector< T > >
#define LIM( T ) numeric_limits< T >

template < typename T > inline istream& operator>>( istream &s, vector< T > &v ){ for ( T &t : v ) { s >> t; } return s; }
template < typename T > inline ostream& operator<<( ostream &s, const vector< T > &v ){ for ( int i = 0; i < int( v.size() ); ++i ){ s << ( " " + !i ) << v[i]; } return s; }
template < typename T > inline T fromString( const string &s ) { T res; istringstream iss( s ); iss >> res; return res; };
template < typename T > inline string toString( const T &a ) { ostringstream oss; oss << a; return oss.str(); };

#define REP2( i, n ) REP3( i, 0, n )
#define REP3( i, m, n ) for ( int i = ( int )( m ); i < ( int )( n ); ++i )
#define GET_REP( a, b, c, F, ... ) F
#define REP( ... ) GET_REP( __VA_ARGS__, REP3, REP2 )( __VA_ARGS__ )
#define FOR( e, c ) for ( auto &e : c )
#define ALL( c ) ( c ).begin(), ( c ).end()
#define AALL( a, t ) ( t* )a, ( t* )a + sizeof( a ) / sizeof( t )
#define DRANGE( c, p ) ( c ).begin(), ( c ).begin() + p, ( c ).end()

#define SZ( v ) ( (int)( v ).size() )
#define PB push_back
#define EM emplace
#define EB emplace_back
#define BI back_inserter

#define EXIST( c, e ) ( ( c ).find( e ) != ( c ).end() )

#define MP make_pair
#define fst first
#define snd second

#define DUMP( x ) cerr << #x << " = " << ( x ) << endl

constexpr int MOD = 1000000007;

int main()
{
	cin.tie( 0 );
	ios::sync_with_stdio( false );

	int N, L;
	cin >> N >> L;

	VI xs( N ), as( N );
	cin >> xs >> as;

	VVI dp( L + 1, VI( N + 1, 0 ) );
	dp[0][0] = 1;

	REP( i, L )
	{
		REP( j, N + 1 )
		{
			( dp[ i + 1 ][j] += dp[i][j] ) %= MOD;
			if ( j < N && xs[j] <= i && ( xs[j] == i || as[j] && ( i - xs[j] ) % as[j] == 0 ) )
			{
				( dp[ i + 1 ][ j + 1 ] += dp[i][j] ) %= MOD;
			}
		}
	}

	cout << dp[L][N] << endl;

	return 0;
}