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 Practice
{
    class Program
    {
        static void Main(string[] args)
        {
            Solve();
        }
        private const long MOD = 1000000007;

        static void Solve()
        {
            var o = ReadAndParseIntArr();
            long n = o[0];
            long t = o[1];
            var ts = ReadAndParseLongArr();

            long ans = t;// n番目
            for(long i=n-2;i>=0;--i)
            {
                if( ts[i+1] - ts[i] >= t)
                {
                    ans += t;
                }
                else
                {
                    ans += ts[i + 1] - ts[i];
                }
            }
            Console.WriteLine(ans);
        }

        #region よく使う便利関数
        private static bool isPrime(long x)
        {
            if (x == 2) { return true; }
            if (x < 2 || x % 2 == 0) { return false; }
            long i = 3;
            while (i * i <= x)
            {
                if (x % i == 0) { return false; }
                i = i + 2;
            }
            return true;
        }
        private static long lcm(long x, long y)
        {
            return x / gcd(x, y) * y;
        }
        private static long gcd(long x, long y)
        {
            return y > 0 ? gcd(y, x % y) : x;
        }
        private static long pow(long x, long n)
        {
            if (n == 0) { return 1; }
            long res = pow(x * x % MOD, n / 2);
            if (n % 2 == 1)
            {
                res = res * x % MOD;
            }
            return res;
        }
        private static int ReadAndParseInt()
        {
            return int.Parse(Console.ReadLine());
        }
        private static int[] ReadAndParseIntArr()
        {
            return Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
        }
        private static long ReadAndParseLong()
        {
            return long.Parse(Console.ReadLine());
        }
        private static long[] ReadAndParseLongArr()
        {
            return Array.ConvertAll(Console.ReadLine().Split(' '), long.Parse);
        }
        #endregion
    }
}