Problem:
Score : 300 points

Problem Statement

In a public bath, there is a shower which emits water for T seconds when the switch is pushed.

If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds.
Note that it does not mean that the shower emits water for T additional seconds.

N people will push the switch while passing by the shower.
The i-th person will push the switch t_i seconds after the first person pushes it.

How long will the shower emit water in total?

Constraints

1 ≤ N ≤ 200,000

1 ≤ T ≤ 10^9

0 = t_1 < t_2 < t_3 < , ..., < t_{N-1} < t_N ≤ 10^9

T and each t_i are integers.

Input

Input is given from Standard Input in the following format:

N T
t_1 t_2 ... t_N

Output

Assume that the shower will emit water for a total of X seconds. Print X.

Sample Input 1

2 4
0 3

Sample Output 1

7

Three seconds after the first person pushes the water, the switch is pushed again and the shower emits water for four more seconds, for a total of seven seconds.

Sample Input 2

2 4
0 5

Sample Output 2

8

One second after the shower stops emission of water triggered by the first person, the switch is pushed again.

Sample Input 3

4 1000000000
0 1000 1000000 1000000000

Sample Output 3

2000000000

Sample Input 4

1 1
0

Sample Output 4

1

Sample Input 5

9 10
0 3 5 7 100 110 200 300 311

Sample Output 5

67

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

{-# LANGUAGE BangPatterns #-}
import Data.List
import Data.Maybe
import Control.Monad
import Data.Function (fix)
import Data.Array (Array, array, (!))
import qualified Data.ByteString as BS
import qualified Data.ByteString.Char8 as BS8
 
getInts :: IO [Int]
getInts = (map parseInt . BS8.words) <$> BS.getLine 
    where parseInt = fst . fromJust . BS8.readInt

data Tree a = Tree (Tree a) a (Tree a)

instance Functor Tree where
    fmap f (Tree l m r) = Tree (fmap f l) (f m) (fmap f r)

index :: Tree a -> Int -> a
index (Tree _ m _) 0 = m
index (Tree l _ r) n = case (n - 1) `divMod` 2 of
    (q,0) -> index l q
    (q,1) -> index r q

nats :: Tree Int
nats = go 0 1
    where go !n !s = Tree (go l s') n (go r s')
            where l = n + s
                  r = l + s
                  s' = s * 2

memo :: (Int -> a) -> (Int -> a)
memo f = index cached
    where cached = fmap f nats

memo2 :: (Int -> Int -> a) -> (Int -> Int -> a)
memo2 f = memo (\x -> memo (f x))

memo3 :: (Int -> Int -> Int -> a) -> (Int -> Int -> Int -> a)
memo3 f = memo (\x -> memo2 (f x))

solve :: Int -> [Int] -> Int
solve t = fst . foldl (\(a,b) x -> (a + min t (x+t-b), x+t)) (0,0)

main :: IO ()
main = do
    [_, t] <- getInts
    arr <- getInts
    print $ solve t arr