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 Rust (1.15.1) (judge verdict: Accepted)

use std::str::FromStr;

fn get_one<T : FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    return s.trim().parse().ok().unwrap();
}

fn get_vec<T : FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    return s.trim().split_whitespace().map(|e| e.parse().ok().unwrap()).collect();
}

fn main() {
    let nt : Vec<i32> = get_vec();
    let xs : Vec<i32> = get_vec();
    let n = nt[0] as usize;
    let t = nt[1];
    let mut ans = 0;

    for i in 1..n {
        if xs[i] < xs[i-1] + t {
            ans += xs[i] - xs[i-1];
        }
        else {
            ans += t;
        }
    }
    ans += t;
    println!("{}", ans);
}