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<vector>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define pb push_back
#define gt getchar()
#define ll long long
#define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
typedef std::pair<int,int> P;
#define mk std::make_pair
#define fr first
#define sc second
inline int in()
{
	int k=0;char ch=gt;bool p=1;
	while(ch<'-')ch=gt;if(ch=='-')ch=gt,p=0;
	while(ch>'-')k=k*10+ch-'0',ch=gt;
	return p?k:-k;
}
const int N=4e5+5;
ll ans=1e18;
int x[N],y[N],n,Mx,Mi,mx,mi,fz[N<<1],tot;
std::vector<int>G[N<<1];
inline void upd(int x,int i){if(x>Mx)Mx=x,mx=i;if(x<Mi)Mi=x,mi=i;}
inline void work1()
{
	int mmx=-2e9,mmi=2e9;
	mmx=std::max(mmx,y[mx]);
	mmi=std::min(mmi,x[mi]);
	for(int i=1;i<=n;++i)
		if(i!=mx&&i!=mi)
			mmx=std::max(mmx,y[i]),mmi=std::min(mmi,x[i]);
	ans=std::min(ans,1ll*(mmx-Mi)*(Mx-mmi));
}
inline void work2()
{
	for(int i=1;i<=n;++i)fz[++tot]=x[i],fz[++tot]=y[i];
	std::sort(fz+1,fz+tot+1);tot=std::unique(fz+1,fz+tot+1)-fz-1;
	int mmi=std::min(y[mx],x[mi]),mmx=y[mx]+x[mi]-mmi;
	for(int i=1;i<=n;++i)if(i!=mx&&i!=mi)mmi=std::min(mmi,x[i]),mmx=std::max(mmx,y[i]);
	int l=std::lower_bound(fz+1,fz+tot+1,mmi)-fz;
	for(int i=1;i<=n;++i)if(i!=mx&&i!=mi)G[std::lower_bound(fz+1,fz+tot+1,y[i])-fz].pb(i);
	for(int i=2;i<=l;++i)
	{
		ans=std::min(ans,1ll*(Mx-Mi)*(mmx-fz[i]));
		for(int j:G[i])mmx=std::max(mmx,x[j]);
	}
}
int main()
{
	n=in();Mx=-2e9,Mi=2e9;
	for(int i=1;i<=n;++i)
	{
		upd(x[i]=in(),i),upd(y[i]=in(),i);
		if(x[i]<y[i])std::swap(x[i],y[i]);
	}
	work1(),work2();printf("%lld\n",ans);
	return 0;
}