Problem:
Score : 300 points

Problem Statement

There is an integer sequence A of length N whose values are unknown.

Given is an integer sequence B of length N-1 which is known to satisfy the following:

B_i \geq \max(A_i, A_{i+1})

Find the maximum possible sum of the elements of A.

Constraints

All values in input are integers.

2 \leq N \leq 100

0 \leq B_i \leq 10^5

Input

Input is given from Standard Input in the following format:

N
B_1 B_2 ... B_{N-1}

Output

Print the maximum possible sum of the elements of A.

Sample Input 1

3
2 5

Sample Output 1

9

A can be, for example, ( 2 , 1 , 5 ), ( -1 , -2 , -3 ), or ( 2 , 2 , 5 ). Among those candidates, A = ( 2 , 2 , 5 ) has the maximum possible sum.

Sample Input 2

2
3

Sample Output 2

6

Sample Input 3

6
0 153 10 10 23

Sample Output 3

53

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
  Motto - try hard to become an expert
*/
#include <bits/stdc++.h>
#include <map>
#include <algorithm>
#include <vector>
#include <string.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <chrono>
#include <thread>
#include <iomanip>
 
//define
#define rep(i,n)    for(int i=0;i<n;i++)
#define repu(i,k,n) for(int i=k;i<=n;i++)
#define repd(i,k,n) for(int i=k;i>=n;i--)
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define INF (int)INT_MAX
#define all(v)  v.begin(), v.end()
#define endl "\n"
 
//typedef
typedef long long int ll;
typedef long double ld;
typedef std::pair<int, int> pii;
typedef std::vector<int> vi;
typedef std::vector< pii > vii;
typedef std::vector< ll > vll;
typedef std::vector< std::pair < pii, int > > vpp;
 
const int N = 3*1e5 +5 ;
const long long MOD = 998244353;
const long long inf = 1e18;
 
using namespace std;
 
ll MAXN=100001;
ll spf[100001]; 
 
ll modpow(ll x,ll n){
    ll res=1;
    while(n>0){
        if(n&1) res=res*x%MOD;
        x=x*x%MOD;
        n>>=1;
    }
    return res;
}
 
ll modFact(ll n, ll p) 
{ 
    if (n >= p) 
        return 0; 
  
    ll result = 1; 
    for (ll i = 1; i <= n; i++) 
        result = (result * i) % p; 
  
    return result;  
} 
 
bool isPrime(int x)
{
    if(x%2==0)
        return 0;
    for(int i=3;i<=sqrt(x);i+=2)
    {
        if(x%i==0)
            return 0;
    }
    return 1;
}
 

 
void solve()
{
	int n  ; 
	cin>>n ; 
	int b[n-1]   , a[n] ; 
	rep(i , n-1) cin>>b[i] ; 
	a[0] = b[0] ; 
	a[n-1] = b[n-2] ;
	repu(i , 1 , n-2){
		a[i]  = min(b[i-1], b[i]) ;
	}
	int ans =0 ;
	rep(i , n){
		ans +=a[i] ;
	}
	cout<<ans<<endl;
}
 
int main() {
    // #ifndef ONLINE_JUDGE
    //     // for getting input from input.txt
    //     freopen("input.txt", "r", stdin);
    //     // for writing output to output.txt
    //     freopen("output.txt", "w", stdout);
    // #endif
    
    // ios_base::sync_with_stdio(0);
    // cin.tie(NULL);
    int t = 1;
    // cin>>t;
    while(t--)
    {
        solve();
    } 
    // cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}