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)

    /*
    Life Before Death, Strength Before Weakness, Journey Before Destination
    ------------------------------------------------------------------------
    You don't become great by trying to be great.
    You become great by wanting to do something,
    and then doing it so hard that you become great in the process.
    */
 
    //#pragma comment(linker, ack:200000000")
    #pragma GCC optimize ("Ofast")
    #pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
    #include <bits/stdc++.h>
    using namespace std;
    
    #define TRACE
    
    #ifdef TRACE
    #define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <typename Arg1>
    void __f(const char* name, Arg1&& arg1){
      cerr << name << " : " << arg1 << std::endl;
    }
    template <typename Arg1, typename... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args){
      const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
    }
    #else
    #define trace(...)
    #endif
    
    #define rep(i, n)    for(ll i = 0; i < (n); ++i)
    #define repA(i, a, n)  for(ll i = a; i <= (n); ++i)
    #define repD(i, a, n)  for(ll i = a; i >= (n); --i)
    #define trav(a, x) for(auto& a : x)
    #define all(x) x.begin(), x.end()
    #define sz(x) (int)(x).size()
    #define fill(a)  memset(a, 0, sizeof (a))
    #define fst first
    #define snd second
    //#define mp make_pair
    #define pb push_back
    #define PI           3.141592653589793238462643383279502884197169399  /* pi */
    // #define te    
    typedef long double ld;
    typedef long long ll;
    typedef pair<ll, ll> pii;
    typedef vector<ll> vi;
    const int maxN = 100100;

    void solve() {
        ll n;
        cin >> n;
        ll a[n], sum = 0;
        rep(i, n-1) cin >> a[i];
        sum = a[0] + a[n-2];
        rep(i, n-2) sum+=min(a[i], a[i+1]);
        cout << sum << endl;
    }
 
    int main() {
      cin.sync_with_stdio(0); cin.tie(0);
      cin.exceptions(cin.failbit);
      // ll ti;   cin >> ti;  while(ti--)
      solve();
      return 0;
    }