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;

#define REP(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define RREP(i,n) for(int (i)=(int)(n)-1;i>=0;i--)
#define REMOVE(Itr,n) (Itr).erase(remove((Itr).begin(),(Itr).end(),n),(Itr).end())
#define MOD  1000000007
#define INF  0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
typedef long long ll;

ll N,T;

int main() {
    
    cin >> N >> T;
    vector<ll> t(N);
    REP(i,N) cin >> t[i];
    t.push_back(1e18);
    
    ll ans = 0;
    ll time = t[0];
    for(int i=1;i<t.size();i++) {
        if(t[i] - time <= T) {
            ans += t[i] - time;
        } else {
            ans += T;
        }
        time = t[i];
    }
    
    cout << ans << endl;
    
    return 0;
}