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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
int n, m;
int a[1005][1005];
int main()
{
	scanf("%d %d",&n,&m);
	for(int i = 1;i <= n;i++)
	{
		for(int j = 1;j <= m;j++)
		{
			scanf("%d",&a[i][j]);
		}
	}
	int f = 0;
	for(int i = 1;i <= n;i++)
	{
		for(int j = 1;j <= m;j++)
		{
			if(a[i][j] == 0)
			{
			//	printf("%d %dmmmm\n",i, j);
				for(int l = j;l >= 1;l--)
				{
					if(a[i][l] == 1)
					{
						f++;
					//	printf("l\n");
						break;
					}
				}
				for(int r = j;r <= m;r++)
				{
					if(a[i][r] == 1)
					{
						f++;
					//	printf("r\n");
						break;
					}
				}
				for(int u = i;u >= 1;u--)
				{
					if(a[u][j] == 1)
					{
						f++;
					//	printf("u\n");
						break;
					}
				}
				for(int d = i;d <= n;d++)
				{
					if(a[d][j] == 1)
					{
						f++;
					//	printf("d\n");
						break;
					}
				}
			}
		}
	}
	printf("%d\n",f);
	return 0;
}