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 Ruby (2.3.3) (judge verdict: Accepted)

a = gets.chomp.split(" ").map(&:to_i)
b = gets.chomp.split(" ").map(&:to_i)
sum = 0
#sum += a[1]
kouho = a[1]
strt=0

for i in 0..a[0]-1
  if b[i] > kouho
    sum += kouho-strt
    strt = b[i]
    kouho = b[i]+ a[1] 
  else
    kouho = b[i]+ a[1] 
  end
  #p kouho
  #if a[1]-before-b[i-1] > 0
  #  p a[1]-before-b[i-1]
  #  sum -= (a[1]-before - b[i-1]) 
  #else
  #  before = b[i]
  #end
end
p sum+kouho-strt