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<bits/stdc++.h>
#include<iomanip>
using namespace std;
typedef long long ll;
typedef long  l;
typedef pair<int,int> P;
#define rep(i,n) for(int i=0;i<n;i++)
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
const int INF=1000000001;
const double PI=3.141592653589;
const ll LMAX=1000000000000001;
ll gcd(ll a,ll b){if(a<b)swap(a,b);while((a%b)!=0){a=b;b=a%b;}return b;}
int dx[]={-1,0,1,0};
int dy[]={0,1,0,-1};



int main(){
  ll n,T; cin>>n>>T;
  ll sum=0;
  vector<ll> t(n);
  rep(i,n) cin>>t[i];

  rep(i,n-1){
    if(t[i+1]-t[i]<T) sum+=t[i+1]-t[i];
    else sum+=T;
  }
  sum+=T;
  cout<<sum<<endl;

  return 0;
}