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 Ruby (2.3.3) (judge verdict: Wrong Answer)


N = gets.chomp.split(" ").map(&:to_i)
a = gets.chomp.split(" ").map(&:to_i)
b = gets.chomp.split(" ").map(&:to_i)

REV=1000000007

nb = b#.uniq

HSH = {}

s = a.min
k = b.max

[*s..k].each{|t|
  s=[1] 
  N[1].times do |i|
    s << s.last * t 
  end
  HSH[t] = s
}
=begin
def dig(a, num,id)
  return num if id+1 == N[0]
  a[0].each do |t|
    p dig(a[1..-1], (num*t)%REV,id+1)
  end
end

p dig(a,1,0)
=end

#p a
#a.each do |column|
#  column.each do |row|
 #   p row
#  end
#end

#=begin
HASH = {}

def f(b,n1)
ans = 0
b.repeated_combination(n1) do |k|
  stub = 1
  s = k.group_by{|i| i}
  if HASH[s]
  stub =HASH[s]
  else
  s.each do |key,v|
    stub = (stub * HSH[key][v.size] ) % REV
  end
  HASH[s] = stub
  end
  #stub = 1
  #k.each do |t|
  #  #e = f()
   # stub = (stub * t) % REV
  #end
  ans = (ans + stub) % REV
end

return ans
end


p f(nb,N[1])
#=end