12:17 (UTC): The sample input 1 and 2 were swapped. The error is now fixed. We are very sorry for your inconvenience.

There are N children in AtCoder Kindergarten, conveniently numbered 1 through N. Mr. Evi will distribute C indistinguishable candies to the children.

If child i is given a candies, the child's happiness will become x_i^a, where x_i is the child's excitement level. The activity level of the kindergarten is the product of the happiness of all the N children.

For each possible way to distribute C candies to the children by giving zero or more candies to each child, calculate the activity level of the kindergarten. Then, calculate the sum over all possible way to distribute C candies. This sum can be seen as a function of the children's excitement levels x_1,..,x_N, thus we call it f(x_1,..,x_N).

You are given integers A_i,B_i (1≦i≦N). Find <image> modulo 10^9+7.

Constraints

* 1≦N≦400
* 1≦C≦400
* 1≦A_i≦B_i≦400 (1≦i≦N)

Input

The input is given from Standard Input in the following format:


N C
A_1 A_2 ... A_N
B_1 B_2 ... B_N


Output

Print the value of <image> modulo 10^9+7.

Examples

Input

2 3
1 1
1 1


Output

4


Input

1 2
1
3


Output

14


Input

2 3
1 1
2 2


Output

66


Input

4 8
3 1 4 1
3 1 4 1


Output

421749


Input

3 100
7 6 5
9 9 9


Output

139123417
#include <cstdio>
#include <cstring>
#include <algorithm>
#define R register
using namespace std;
const int mod=1000000007;
const int MAXN=410;
int a[MAXN],b[MAXN],po[MAXN][MAXN];
int dp[MAXN][MAXN];

int add(int x,int y)
{return (x+y)%mod;}
int sub(int x,int y)
{return (x-y+mod)%mod;}
int mul(int x,int y)
{return 1LL*x*y%mod;}

int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for(R int i=1;i<=n;++i)scanf("%d",&a[i]);
	for(R int i=1;i<=n;++i)scanf("%d",&b[i]);
	for(R int i=1;i<=400;++i)
	{
		int t=1;
		po[i][0]=i;
		for(R int j=1;j<=m;++j)
		{
			t=mul(t,i);
			po[i][j]=add(po[i-1][j],t);
		}
	}
	dp[0][0]=1;
	for(R int i=1;i<=n;++i)
	  for(R int j=0;j<=m;++j)
		for(R int k=0;k<=m;++k)
		{
			if(j+k>m)break;
			dp[i][k+j]=add(dp[i][k+j],mul(dp[i-1][k],sub(po[b[i]][j],po[a[i]-1][j])));
		}
	printf("%d\n",dp[n][m]);
	return 0;
}