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: Accepted)

#include <bits/stdc++.h>

#define ll long long
#define FOR(i,a,b) for (int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define debug(x) cerr<<#x<<": "<<x<<'\n'

using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;

int dy[]={0, 0, 1, -1, 0};
int dx[]={1, -1, 0, 0, 0};

template<class T=int>
T in(){T x;cin>>x;return (x);}
template<class T>
void print(T& x){cout<<x<<'\n';}

const int MOD =(int)1e9+7;
const int MAX =510000;

ll fac[MAX],finv[MAX],inv[MAX];
void COMint(){
    fac[0]=fac[1]=1;
    finv[0]=finv[1]=1;
    inv[1]=1;
    for(int i=2;i<MAX;i++){
        fac[i]=fac[i-1]*i%MOD;
        inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
        finv[i]=finv[i-1]*inv[i]%MOD;
    }
}
ll COM(int n,int k){
    if(n<k) return 0;
    if(n<0||k<0)return 0;
    return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;
}

ll gcd(ll a,ll b){
    if(b==0)return a;
    if(a>b){
        swap(a,b);
    }
    return gcd(a,b%a);
}
ll lcm(ll a,ll b){
    ll g;g=gcd(a,b);
    return a*b/g;
}
int main() {
    //COMint();
    ll N=in();
    ll T=in();
    ll t[N];
    ll ans;ans=0;
    REP(i,N){
        cin>>t[i];
    }
    ll endyu;endyu=T;
    for(int i=1;i<N;i++){
        if(t[i]>t[i-1]+T){
            ans+=T;
        }else{
            ans+=t[i]-t[i-1];
        }
    }
    ans+=T;
    print(ans);
}