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 <iostream>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
typedef struct pair<int,int> P;
typedef long long ll;

ll n,k,l;

int dp1[16002][3]={};
int dp2[16002][3]={};
ll c[200001]={};
int main() {
	int n;
	ll t;
	ll t1;
	ll s=0;

	cin>>n>>t;
	ll e=0;
	for(int i=0;i<n;++i){
		cin>>t1;
		if(t1<e){
			s=s+t-e+t1;
		}else{
			s+=t;
		}
		e=t1+t;
	}
	cout<<s<<endl;
//	int ls=s.length();
//
//	int m=0;
//	int a1=0;
//	int a2=0;
//	int a3=0;
//	int a0=0;
//	bool flg=true;
//	for(a1=1;(s[a1]==s[a1-1])&&(a1<ls);a1++);
//	m=a1-a0;
//	for(a2=a1+1;(s[a2]==s[a2-1])&&(a2<ls);a2++);
//	m=max(a2-a1,a1-a0);
//	if(a2==ls)flg=false;
//	for(a3=a2+1;a3<ls;a3++){
//		for(;(s[a3]==s[a3-1])&&(a3<ls);a3++);
//		m=max(m,((a2-a1)+min(a1-a0,a3-a2)));
//		a0=a1;
//		a1=a2;
//		a2=a3;
//	}
//	if(flg)m=max(m,((a2-a1)+min(a1-a0,a3-a2)));
//	cout<<m<<endl;
	return 0;
}