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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace cswork
{
    class Program
    {
        class Scanner
        {
            string[] s;
            int i;

            char[] cs = new char[] { ' ' };

            public Scanner()
            {
                s = new string[0];
                i = 0;
            }

            public string next()
            {
                if (i < s.Length) return s[i++];
                string st = Console.ReadLine();
                while (st == "") st = Console.ReadLine();

                s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
                i = 0;
                return next();
            }

            public int nextInt()
            {
                return int.Parse(next());
            }

            public long nextLong()
            {
                return long.Parse(next());
            }

            public double nextDouble()
            {
                return double.Parse(next());
            }
        }

        public Program() { }

        static void Main(string[] args)
        {
            new Program().Omame();
        }

        Scanner cin = new Scanner();
        const int MAXIN = 1024;
        bool[] used = new bool[MAXIN];
        int[] perm = new int[MAXIN];

        List<int[]> permList = new List<int[]>();

        public void Permutation(int pos, int n, int length)
        {
            if (pos == n)
            {
                permList.Add((int[])perm.Clone());
            }

            for (int i = 0; i < n; i++)
            {
                if (!used[i])
                {
                    perm[pos] = i;
                    used[i] = true;
                    Permutation(pos + 1, n, length);
                    used[i] = false;
                }
            }
        }

        public static List<long> GeneratePrime(long max)
        {
            long prime;
            double sqrtMax = Math.Sqrt(max);
            var primeList = new List<long>();

            List<long> searchList = new List<long>();
            for (long i = 2; i <= max - 1; i++)
            {
                searchList.Add(i);
            }

            do
            {
                prime = searchList.First();
                primeList.Add(prime);
                searchList.RemoveAll(n => n % prime == 0);
            } while (prime < sqrtMax);

            primeList.AddRange(searchList);

            return primeList;
        }

        // 最小公倍数
        public static long Lcm(long a, long b)
        {
            return a * b / Gcd(a, b);
        }

        // ユークリッドの互除法 
        public static long Gcd(long a, long b)
        {
            if (a < b)
                return Gcd(b, a);
            while (b != 0)
            {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }

        public static readonly string Yes = "Yes";
        public static readonly string YES = "YES";
        public static readonly string No = "No";
        public static readonly string NO = "NO";
        public static readonly string OK = "OK";
        public static readonly string NG = "NG";

        void Omame()
        {
            long N = cin.nextLong();
            long T = cin.nextLong();

            if(N == 1)
            {
                Console.WriteLine(T);
                return;
            }
            long now = cin.nextLong();
            long ans = 0;
            for (int i = 1; i < N; i++)
            {
                long next = cin.nextLong();

                if(next  >= now + T)
                {
                    ans += T;
                }
                else
                {
                    ans += next - now;
                }

                if(i == N - 1)
                {
                    ans += T;
                }
                now = next;
            }
            Console.WriteLine(ans);
        }
    }
}