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++ (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200500;
typedef long long ll;
ll a[maxn];
int main()
{
    int n,t;
    cin>>n>>t;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    int start=a[0];
    ll ans=0;
    ll now=a[0];
   /* for(int i=1;i<n;i++)
    {
        now+=a[i];
        cout << now <<endl;
        if(start+t<=now)
        {
            ans+=t;
            start=now;
        }
        else if(start+t>now)
        {
            ans+=now-start;
            cout << now-start <<endl;
            start=now;
        }
        cout << ans <<endl;
    }*/
    for(int i=1;i<n;i++)
    {
        if(a[i-1]+t<=a[i])
            ans+=t;
        else
            ans+=a[i]-a[i-1];
    }
    ans+=t;
    cout << ans <<endl;
    return 0;
}