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 ALL(obj) (obj).begin(), (obj).end()

#define endl "\n"
#define pb push_back
#define elif else if
#define cout_db cout << fixed

#define MOD 1000000007
#define PI 3.1415926
 
const int dx[]={1,0,-1,0}, dy[]={0,-1,0,1};
const int dx_8[]={1,1,0,-1,-1,-1,0,1}, dy_8[]={0,-1,-1,-1,0,1,1,1};

typedef long long ll;

template <typename T>
inline T GCD(T a,T b){T c;while(b!=0){c=a%b;a=b;b=c;}return a;}
template <typename T>
inline T LCM(T a,T b){T c=GCD(a,b);a/=c;return a*b;}
template <typename T>
inline T nCr(T a,T b){T i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
template <typename T>
inline T nHr(T a,T b){return nCr(a+b-1,b);}
template <typename T>
inline T POW(T a,T b){int i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}

signed main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N, T, t[200000];
	cin >> N >> T;
	for (int i = 0; i < N; i++) {
		cin >> t[i];
	}

	int ans = 0;
	for (int i = 1; i < N; i++) {
		int ti = t[i-1] + T;
		ans += min(ti, t[i]) - t[i-1];
	}

	cout << ans + T << endl;

	return 0;
}