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: Wrong Answer)

#include <stdio.h>
#include <string.h>


int main(void)
{
//	int  i, n;
	unsigned long i, t, sum=0, total, n;
	unsigned long tt[200000];
	
	scanf("%lu %lu", &n, &t );
	for(i=0; i<n; i++){
		scanf(" %lu", &tt[i]);
	}
	
	sum = t;
	for(i=1; i<n; i++){
		//sum = sum + ( (t + tt[i]) - t );
		if(tt[i] > sum){
			sum = sum + t;
			//printf("a\n");
		}else{
			sum = sum + ( ( tt[i] + t ) - sum );
			//printf("tt[%ld]->%ld\n", i, tt[i]);
			//printf("b\n");		
		}
		
	}
	printf("%lu\n", sum );
	
	//getchar();
	//getchar();
	return 0;
}