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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {

    static void solve() {
        int n = nextInt();
        int t = nextInt();
        int[] tt = nextIntArray(n);

        long ans = t;
        for (int i = 0; i < n - 1; i++) {
            ans += Math.min(tt[i + 1] - tt[i], t);
        }

        out.println(ans);
    }

    static final int MOD = 1_000_000_007;
    static long[] fac, finv, inv;

    // 階乗(n!)
    static long factorial(long n) {
        long ans = 1;
        for (long i = n; i > 0; i--) {
            ans = ans * i % MOD;
        }
        return ans;
    }

    // nCkの初期化
    static void comInit(int max) {
        fac = new long[max];
        finv = new long[max];
        inv = new long[max];
        fac[0] = fac[1] = 1;
        finv[0] = finv[1] = 1;
        inv[1] = 1;

        for (int i = 2; i < max; i++) {
            fac[i] = fac[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
    }

    // nCkの計算
    static long com(int n, int k) {
        if (n < k)
            return 0;
        if (n < 0 || k < 0)
            return 0;
        return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
    }

    static PrintWriter out;
    static Scanner sc;

    static int[][] newIntArray(int h, int w, int value) {
        int[][] ret = new int[h][w];
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                ret[i][j] = value;
            }
        }
        return ret;
    }

    static int nextInt() {
        return Integer.parseInt(sc.next());
    }

    static long nextLong() {
        return Long.parseLong(sc.next());
    }

    static String nextString() {
        return sc.next();
    }

    static int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    static List<Integer> nextIntList(int n) {
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(nextInt());
        }
        return list;
    }

    static List<Double> nextDoubleList(int n) {
        List<Double> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add((double) nextInt());
        }
        return list;
    }

    static List<Long> nextLongList(int n) {
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(nextLong());
        }
        return list;
    }

    static char[][] nextCharArray(int h, int w) {
        char[][] c = new char[h][w];
        for (int i = 0; i < h; i++) {
            String str = nextString();
            for (int j = 0; j < w; j++) {
                c[i][j] = str.charAt(j);
            }
        }
        return c;
    }

    static <T extends Comparable<? super T>> void sort(List<T> list) {
        Collections.sort(list);
    }

    // greatest common divisor
    // 最大公約数
    static long gcd(long a, long b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    // least common multiple
    // 最小公倍数
    static long lcm(long a, long b) {
        if (a >= b && a % b == 0)
            return a;
        if (b > a && b % a == 0)
            return b;

        if (a > b) {
            return (a / gcd(a, b)) * b;
        } else {
            return (b / gcd(a, b)) * a;
        }
        // return a * b / gcd(a, b);
    }

    // baseのn乗を計算を返す
    static int pow(int base, int n) {
        int ret = 1;
        for (int i = 0; i < n; i++) {
            ret *= base;
        }
        return ret;
    }

    // return n^k mod m
    static long powMod(long n, long k, long m) {
        if (k == 0) {
            return 1;
        } else if (k % 2 == 1) {
            return powMod(n, k - 1, m) * n % m;
        } else {
            long tmp = powMod(n, k / 2, m);
            return tmp * tmp % m;
        }
    }

    // intをlength桁のbit文字列に変換
    static String toBitString(int length, int n) {
        StringBuilder sb = new StringBuilder();
        for (int i = length - 1; i >= 0; i--) {
            if ((n >> i) % 2 == 1) {
                sb.append("1");
            } else {
                sb.append("0");
            }
        }
        return sb.toString();
    }

    public static void main(String[] args) {
        out = new PrintWriter(System.out);
        sc = new Scanner(System.in);

        solve();

        out.flush();
        sc.close();
    }

}