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<algorithm>//sort(all(変数),greater<型名>())で降順に、*max_element(all(変数))で最大値を獲得
#include<bits/stdc++.h>
#include<cmath>//切り上げceil(値)
#include<deque>
#include<iomanip>//setprecision(数字)
#include<iostream>
#include<list>
#include<map>
#include<numeric>
// xとyのgcdは__gcd(x,y),xとyのlcmは(x*y)/__gcd、accumulate(all(変数),0)で要素の全加算
#include<queue>//priority_queue<型名> 変数
#include<set>
#include<stack>
#include<string>//変数.substr(どこから,いくつ)
#include<vector>
#define rep(i,n)  for(ll i=0; i<n; i++)
#define rrep(i,n) for(ll i=n-1; 0<=i; i--)
#define vll vector<ll>
#define vi vector<int>
#define all(x) (x).begin(),(x).end()
#define INF 1e9
using ll = long long;
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

int main(){
  int n,T; cin >> n >> T;
  vi t(n); rep(i,n) cin >> t[i];
  ll ans = 0;
  rep(i,n-1){
    ans += min(T,t[i+1]-t[i]);
  }
  cout << ans+T;
  cout << "\n";
  return 0;
}