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;
typedef long long ll;
typedef long double ld;
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REPR(i,n) for(int i=n; i>-1; --i)
#define ALL(a) (a).begin(),(a).end()
#define INIT(a,n,x); REP(i,(n)){ (a)[i]=(x); }
#define CINA(a,n); REP(i,(n)){ cin >> (a)[i]; }
#define INIT2(a,n,m,x); REP(i,(n)){ REP(j,(m)){(a)[i][j]=(x);} }
#define CINA2(a,n,m); REP(i,(n)){ REP(j,(m)){cin >> (a)[i][j];} }
#define Liny "Yes\n"
#define Linn "No\n"
#define LINY "YES\n"
#define LINN "NO\n"

#define umap unordered_map
//cout << setfill('0') << right << setw(4) << 12; // "0012"
int keta(ll x){ if(x<10){return 1;} else{return keta(x/10) + 1;}}
int keta_wa(ll x){ if(x<10){return x;} else{return keta_wa(x/10) + x%10;} }
int ctoi(char c){ return ( (c>='0' && c<='9')? c - '0': 0 );}
ll sum(ll a[],ll N){ return accumulate(a,a+N,0LL);}
ll gcd(ll a,ll b){if(a<b)swap(a,b); return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){if(a<b){swap(a,b);} return a/gcd(a,b)*b;}

int main(){
  ll N,T; cin>>N>>T;
  ll r = 0;
  ll t;
  ll t_pre = 0;
  REP(i,N){
    cin >> t;
    if(t==0){continue;}
    r += min(t-t_pre,T);
    t_pre = t;
  }
  r += T;
  cout << r;
}