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 (GCC 5.4.1) (judge verdict: Accepted)

#define inf 10000000000
#include <stdio.h>

/*int main(void){
    int n,T,sum;
    int t[200000];
    int i;
    int base,crnt,prev;
    scanf("%d %d",&n,&T);
    for(i=0;i<n;i++){
        //t[i]はi番目の人がボタンを押すタイミング
        scanf("%d",&t[i]);
    }
    base=0;
    crnt=0;
    while(1){
        int prev=base;
        //隣接する配列の差がTよりも大きくなるまで
    while((t[crnt]-t[prev]<T)){
        if(crnt==n-1){
            sum+=(t[crnt]-t[base])+T;
            printf("%d\n",sum);
            return 0;
        }
        //最初はcrntのみ
        if(crnt==prev){
            crnt++;
        }else{
        crnt++;
            prev++;
        }
        
    }
        sum+=(t[prev]-t[base])+T;
        base=crnt;
        prev=crnt;
    }
    printf("%d\n",sum);
}
*/
int min(long a,long b){
    if(a<b)return a;
    else return b;
}
int main(void){
    long n,T,sum=0;
    long t[200001];
    int i;
    int base,crnt,prev;
    scanf("%ld %ld",&n,&T);
    for(i=0;i<n;i++){
        //t[i]はi番目の人がボタンを押すタイミング
        scanf("%ld",&t[i]);
    }
    t[n]=inf;
    for(i=0;i<n;i++){
        sum+=min(t[i+1]-t[i],T);
    }
    printf("%ld\n",sum);
}