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>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize ("-ffloat-store") // to restrict undesirable precision
#pragma GCC optimize ("-fno-defer-pop")// to pop argument of function as soon as it returns
#define all(a) a.begin(),a.end()
#define ll long long int
#define ld long double
ll power(ll a,ll b,ll m){ if(b==0) return 1; if(b==1) return a%m; ll t=power(a,b/2,m)%m; t=(t*t)%m; if(b&1) t=((t%m)*(a%m))%m; return t;}
ll modInverse(ll a, ll m) { return power(a, m-2, m); }
#define ps push_back
#define fs first
#define takeline cin.ignore();
#define sc second
#define N 3000005
#define endl "\n"
#define mod 1000000007
//((1.0l)*BIG MULTIPLY MAGIC?)
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//-----------------------------------------------
int main(){
	
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	
	ll i,j,k,l,n,t;
	cin>>n>>t;
	ll ar[n+1];
	for(i=1;i<=n;i++){
		cin>>ar[i];
	}

	ll ans=t;
	for(i=2;i<=n;i++){
		k=ar[i]-ar[i-1];
		if(k>t) ans+=t;
		else ans+=k;
	}

	cout<<ans;
	return 0;
}