Problem:
Score : 800 points

Problem Statement

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 modulo 10^9+7.

Constraints

1≦N≦400

1≦C≦400

1≦A_i≦B_i≦400 (1≦i≦N)

Partial Score

400 points will be awarded for passing the test set satisfying A_i=B_i (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 modulo 10^9+7.

Sample Input 1

2 3
1 1
1 1

Sample Output 1

4

This case is included in the test set for the partial score, since A_i=B_i.
We only have to consider the sum of the activity level of the kindergarten where the excitement level of both child 1 and child 2 are 1 (f(1,1)).

If child 1 is given 0 candy, and child 2 is given 3 candies, the activity level of the kindergarten is 1^0*1^3=1.

If child 1 is given 1 candy, and child 2 is given 2 candies, the activity level of the kindergarten is 1^1*1^2=1.

If child 1 is given 2 candies, and child 2 is given 1 candy, the activity level of the kindergarten is 1^2*1^1=1.

If child 1 is given 3 candies, and child 2 is given 0 candy, the activity level of the kindergarten is 1^3*1^0=1.

Thus, f(1,1)=1+1+1+1=4, and the sum over all f is also 4.

Sample Input 2

1 2
1
3

Sample Output 2

14

Since there is only one child, child 1's happiness itself will be the activity level of the kindergarten. Since the only possible way to distribute 2 candies is to give both candies to child 1, the activity level in this case will become the value of f.

When the excitement level of child 1 is 1, f(1)=1^2=1.

When the excitement level of child 1 is 2, f(2)=2^2=4.

When the excitement level of child 1 is 3, f(3)=3^2=9.

Thus, the answer is 1+4+9=14.

Sample Input 3

2 3
1 1
2 2

Sample Output 3

66

Since it can be seen that f(1,1)=4 , f(1,2)=15 , f(2,1)=15 , f(2,2)=32, the answer is 4+15+15+32=66.

Sample Input 4

4 8
3 1 4 1
3 1 4 1

Sample Output 4

421749

This case is included in the test set for the partial score.

Sample Input 5

3 100
7 6 5
9 9 9

Sample Output 5

139123417

Solution in Python (2.7.6) (judge verdict: Wrong Answer)

import bisect
import sys
import math
import itertools
sys.setrecursionlimit(10000)

INF = float('inf')

# input macro
def i():
    return int(raw_input())
def ii():
    return map(int,raw_input().split(" "))
def s():
    return raw_input()
def ss():
    return raw_input().split(" ")
def slist():
    return list(raw_input())
#

def join(s):
    return ''.join(s)

#iterate macro
def piter(n,m):
    return itertools.permutations(n,m)
def citer(n,m):
    return itertools.combinations(n,m)

#modulo macro
def modc(a,b,m):
    c = 1
    for i in xrange(b):
        c = c * (a - i) % m
        c = c * modinv(i + 1,m) % m
    return c
 
def gcd(a, b):
    (x, lastx) = (0, 1)
    (y, lasty) = (1, 0)
    while b != 0:
        q = a // b
        (a, b) = (b, a % b)
        (x, lastx) = (lastx - q * x, x)
        (y, lasty) = (lasty - q * y, y)
    return (lastx, lasty, a)
 
def modinv(a, m):
    (inv, q, gcd_val) = gcd(a, m)
    return inv % m

#bisect macro
def index(a, x):
    #Locate the leftmost value exactly equal to x
    i = bisect_left(a, x)
    if i != len(a) and a[i] == x:
        return i
    return -1

#memoize macro
def memoize(f):
    cache = {}
    def helper(*args):
        if args not in cache:
            cache[(args)] = f(*args)
        return cache[args]
    return helper

@memoize
def nck(a,b,m):
    b = min([a-b,b])
    if (b>a or b<0 or a<0):
        return 0
    elif a == 0:
        return 1
    return (nck(a-1,b-1,m)+nck(a-1,b,m)) % m

###########
Q=10**9+7
N,C = ii()
A=ii()
B=ii()
@memoize
def f(a,n):
    if n==1:
        return (A[0]**a)%Q
    sum=0
    for i in range(a+1):
        sum += f(a-i,n-1)*(A[n-1]**i)%Q
    return sum

print f(C,N)