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<algorithm>
#include<vector>
using LL = long long;
LL N,w;
LL W, V;
LL size[4];
LL ans;
std::vector<LL> dp1,dp2,dp3,dp4;
void value(LL x, LL y, LL z, LL q) {
	LL sum = 0;
	for (LL aa = 0; aa < x; aa++) {
		sum += dp1[aa];
	}
	for (LL aa = 0; aa < y; aa++) {
		sum += dp2[aa];
	}
	for (LL aa = 0; aa < z; aa++) {
		sum += dp3[aa];
	}
	for (LL aa = 0; aa < q; aa++) {
		sum += dp4[aa];
	}
	ans = std::max(ans, sum);
}
void solve() {
	for (LL i = 0; i <= dp1.size(); i++) {
		for (LL j = 0; j <= dp2.size(); j++) {
			for (LL k = 0;k <= dp3.size(); k++) {
				for (LL l = 0; l <= dp4.size(); l++) {
					LL ww=0;
					ww += i*w;
					ww += j*(w + 1);
					ww += k*(w + 2);
					ww += l*(w + 3);
					if (ww <= W) {
						value(i, j, k, l);
					}
				}
			}
		}
	}
}
int main() {
	std::cin >> N >> W;
	std::cin >> w>> V;
	dp1.push_back(V);
	for (LL i = 1; i < N; i++) {
		LL tmp,Q;
		std::cin >> tmp >> Q;
		if (tmp - w == 0) dp1.push_back(Q);
		if (tmp - w == 1) dp2.push_back(Q);
		if (tmp - w == 2) dp3.push_back(Q);
		if (tmp - w == 3) dp4.push_back(Q);
 
	}
	std::sort(dp1.rbegin(), dp1.rend());
	std::sort(dp2.rbegin(), dp2.rend());
	std::sort(dp3.rbegin(), dp3.rend());
	std::sort(dp4.rbegin(), dp4.rend());
	solve();
	std::cout << ans ;
	return 0;
}