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<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x); i<=(y); i++)
#define ll long long
#define N 405
#define mod 1000000007
using namespace std;
int n,c,a[N],b[N],sum[N][N],f[2][N];
void upd(int &x,int y){ x+=y; if (x>=mod) x-=mod; }
void pre(){
	rep (i,1,N-1){//sum[i][j]=1^j+2^j+3^j+...+i^j;
		sum[i][0]=1;
		rep (j,1,N-1) sum[i][j]=(ll)sum[i][j-1]*i%mod;//i^j
		rep (j,0,N-1) upd(sum[i][j],sum[i-1][j]);
	}
}
int main(){
	scanf("%d%d",&n,&c);
	pre();
	rep (i,1,n) scanf("%d",&a[i]);
	rep (i,1,n) scanf("%d",&b[i]);
	int nk=0,k=1;
	f[0][0]=1;//f[i][j]表示前i个小朋友分了j颗糖
	rep (i,0,n-1){
		k^=1; nk^=1; memset(f[nk],0,sizeof(f[nk]));
		rep (j,0,c) for (int t=0; t+j<=c; t++)//这一轮分了t颗糖
			upd(f[nk][t+j],(ll)f[k][j]*(sum[b[i+1]][t]+mod-sum[a[i+1]-1][t])%mod);
	}
	printf("%d\n",f[nk][c]);
	return 0;
}