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: Wrong Answer)

#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<map>
#include<cstdio>
#include<stack>
#include<queue>
#include<fstream>
#include<cstdio>

using namespace std;
#define ok1 printf("ok1\n");
#define ok2 printf("ok2\n");
#define M 1000000
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=(s);i<(n);i++)
#define repr(i,n) for(int i=n-1;i>=0;i--)
#define REPR(i,s,n) for(int i=(s);i>=(g);(i)--)
#define all(a) (a).begin(),(a).end()
#define reall(a) (a).rbegin(),(a).rend()
#define pb push_back
#define MIN(a,b) a=min((a),(b))
#define MAX(a,b) a=max((a),(b))
#define SIZE(v) (int)v.size()


typedef vector<int> vi;
typedef vector<string> vs;
typedef long long ll;
typedef vector<ll> vll;

#ifdef TESTING
#define DEBUG fprintf(stderr,"====TESTING====\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define DEBUG 
#define VALUE(x)
#define debug(...)
#endif
/*
inline string IntToString(ll a) {
	char x[100];
	sprintf_s(x, "%lld", a); string s = x;
	return s;
}

inline ll StringToInt(string a) {
	char x[100]; ll res;
	strcpy_s(x, a.c_str()); sscanf_s(x, "%lld", &res);
	return res;
}
*/

inline string uppercase(string s) {
	int n = SIZE(s);
	rep(i, n) if (s[i] >= 'a' && s[i] <= 'z') s[i] = s[i] - 'a' + 'A';
	return s;
}

inline string lowercase(string s) {
	int n = SIZE(s);
	rep(i, n) if (s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a';
	return s;
}


int main()
{
	int n, t, ans = 0;
	bool x = true;
	cin >> n >> t;
	vi v(n);
	for (int i = 0; i < n; i++) {
		cin >> v[i];
		if (v[i] == 0)
			ans += t;
		else if (t >= v[i] && n - 1 == i) {
			ans += (t + (v[i] - t));
			//ok1
		}
		else if (t <= v[i] && t > v[i - 1]) {
			ans += t + v[i - 1];
		}
		else if (t <= v[i])
			ans += t;
		else
			continue;
		//cout << ans - t << " " << v[i] << endl;
		
		//cout << ans << " ";
	}
	//cout << endl;
	cout << ans << endl;
	return 0;
}