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: Wrong Answer)

{-# LANGUAGE MultiParamTypeClasses,FlexibleContexts,FlexibleInstances,TypeSynonymInstances,BangPatterns,RankNTypes,TupleSections #-}
import Control.Monad
import Control.Monad.ST
import Control.Applicative
import Control.Arrow
import Debug.Trace
import Text.Printf

import Data.List
import Data.Int
import Data.Bits
import Data.Maybe
import Data.Monoid
import Data.Array.Unboxed
import Data.Array.ST
import Data.Array.Base
import Data.Function(on)
import Data.Tuple(swap)
import qualified Data.Map as M
import qualified Data.Set as S
import qualified Data.ByteString.Char8 as B

main :: IO ()
main = do
    (n,c) <- readIntPair
    as <- readInts
    bs <- readInts
    let arr :: Array (Int,Int) Modulo
        arr = runSTArray $ do
            vis <- newUArray ((0,0),(c,n)) False
            dp <- newArray ((0,0),(c,n)) 0
            let go v _ [] = if v == 0 then return 1 else return 0
                go v i (x:xs) = do
                    b <- readArray vis (v,i)
                    if b then
                        readArray dp (v,i)
                    else do
                        l <- forM [0..v] $ \a -> do
                            r <- go (v - a) (i+1) xs
                            return $ (fromInteger (fromIntegral x))^a * r
                        thru (writeArray dp (v,i)) (sum l)
            go c 0 as
            return dp
    print $ arr ! (c,0)




modulo :: Int64
modulo = 1000000007

newtype Modulo = Modulo { getInteger :: Int64 } deriving(Eq,Ord)
instance Num Modulo where
    Modulo a + Modulo b = Modulo ((a+b) `mod` modulo)
    Modulo a - Modulo b = Modulo ((a-b) `mod` modulo)
    Modulo a * Modulo b = Modulo ((a*b) `mod` modulo)
    abs (Modulo a) = Modulo (abs a)
    signum (Modulo a) = Modulo (signum a)
    fromInteger i = Modulo (fromInteger i)
instance Fractional Modulo where
    recip (Modulo !a) = Modulo (modInv a)
    fromRational = undefined

instance Enum Modulo where
    toEnum x = fromIntegral x
    fromEnum x = fromIntegral $ getInteger x

instance Show Modulo where
    show (Modulo i) = show i
 
modInv i = (modulo + x `mod` modulo) `mod` modulo
    where
    (x,y) = go i modulo
    go !a 0 = (1,0)
    go !a b = let (!y,!x) = go b (a `mod` b) in (x,y - a `div` b * x)


-- IO
readInt :: B.ByteString -> Int
readInt = fromJust . fmap fst . B.readInt
readInts :: IO [Int]
readInts = map readInt . B.words <$> B.getLine
readIntPair :: IO (Int,Int)
readIntPair = l2p . map readInt . take 2 . B.words <$> B.getLine
readAllInts :: IO [[Int]]
readAllInts = map (map readInt . B.words) . B.lines <$> B.getContents
readLns :: Read a => IO [a]
readLns = map read . words <$> getLine

-- Utilities
l2p :: [a] -> (a,a)
l2p (a:b:_) = (a,b)
l2p _ = error "l2p: the input list is too short"
p2l :: (a,a) -> [a]
p2l (a,b) = [a,b]
itof :: Int -> Double
itof = fromIntegral
inf :: Int
inf = maxBound `div` 2

-- Monad
stepM_ :: Monad m => a -> (a -> Bool) -> (a -> a) -> (a -> m ()) -> m ()
stepM_ i0 judge incr step = sub i0 where 
    sub i | judge i = step i >> sub (incr i) 
          | otherwise = return ()
rep_ :: Monad m => Int -> (Int -> m ()) -> m ()
rep_ n = stepM_ 0 (<n) (+1)
thru :: Monad m => (v -> m ()) -> v -> m v
thru m v = m v >> return v
memoize :: (Monad m) => (k -> m (Maybe v)) -> (k -> v -> m ()) -> k -> m v -> m v
memoize getter setter key action = do
    mv <- getter key
    case mv of
        Just v -> return v
        Nothing -> action >>= thru (setter key)

-- Array
newUArray :: (MArray (STUArray s) e (ST s),Ix i) => 
             (i,i) -> e -> ST s (STUArray s i e)
newUArray = newArray
newBArray :: (Ix i) => (i,i) -> e -> ST s (STArray s i e)
newBArray = newArray
defaultArray :: (IArray a e,Ix i) => e -> (i,i) -> [(i,e)] -> a i e
defaultArray = accumArray $ curry snd
fromEdges :: Ix i => (i,i) -> [(i,e)] -> Array i [e]
fromEdges = accumArray (flip (:)) []