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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

import Data.List
import Data.List.Split
import Control.Monad
import qualified Data.Vector as V
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VM
main = mapM_ (print . solve) . chunksOf 3 . map (map read . words) . lines =<< getContents
solve [[n,c],as,bs] = dp VU.! (idx n c)  where
  ps = V.fromList . map VU.fromList . transpose . map f2 . transpose . map f1 $ [0..maximum bs] where
    f1 = scanl (\a b->a*b`mod`m) 1 . replicate c
    f2 = scanl1 (\a b->(a+b)`mod`m)
  qs = V.fromList $ zipWith (\b a->VU.fromList $ map (\k->(p b k-p a k+m)`mod`m) [0..c]) bs $ map pred as
    where p i j = ps V.! i VU.! j
  idx i j = i*(c+1)+j
  dp = VU.create $ do
    dp <- VM.replicate ((c+1)*(n+1)) 0
    VM.write dp (idx 0 0) 1
    forM_ [1..n] $ \i ->
      forM_ [0..c] $ \j ->
        forM_ [0..j] $ \k -> do
          a <- VM.read dp (idx i j)
          b <- VM.read dp (idx (i-1) (j-k))
          VM.write dp (idx i j) ((a+b*(qs V.! (i-1) VU.! k))`mod`m)
    return dp
m = 1000000007