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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stdlib.h>
#include <list>
#include <queue>
#include <map>
#include <stack>
#include <set>
#define inf 0x3f3f3f3f
#define PI acos(double(-1))
using namespace std;

typedef long long ll;
const int maxn=2e5+10;
int n,t[maxn],T;
ll sum;

int main()
{
    scanf("%d%d",&n,&T);
    sum+=T;
    int a;
    int l;
    int a0;
    l=T;
    scanf("%d",&a0);
    for(int i=0;i<n-1;i++)
    {
       // int a;
        scanf("%d",&a);
        if(a>l)
        {
            sum=sum+T;
            l=a+T;
        }
        else{
            sum=sum+a-a0;
            l=a+T;
        }
        a0=a;
    }
    printf("%lld\n",sum);
    return 0;
}