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 <iostream>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;
typedef long long ll;
const int P = 1e9+7, INF = 0x3f3f3f3f;
const int N = 410;
int n, c, a[N], b[N], dp[N][N];
int po[N][N], sum[N][N];

int main() {
	REP(i,1,N-1) { 
		po[i][0] = 1;
		REP(j,1,N-1) po[i][j] = (ll)po[i][j-1]*i%P;
	}
	REP(i,0,N-1) { 
		sum[i][0] = 1;
		REP(j,1,N-1) sum[i][j] = (sum[i][j-1]+po[j][i])%P;
	}
	cin>>n>>c;
	REP(i,1,n) cin>>a[i];
	REP(i,1,n) cin>>b[i];
	dp[0][0] = 1;
	REP(i,1,n) REP(x,0,c) REP(y,0,x) {
		int ret = sum[x-y][b[i]]-sum[x-y][a[i]-1];
		dp[i][x] = (dp[i][x]+(ll)dp[i-1][y]*ret)%P;
	}
	int ans = dp[n][c];
	if (ans<0) ans += P;
	printf("%d\n", ans);
}







