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)

#include <bits/stdc++.h>
#define rep(i, n) REP(i, 0, n)
#define REP(i, l, r) for (int i = l; i < r; ++i)
#define int long long
#ifdef DEBUG
#define dout cout
#else
std::stringstream dout;
#endif
using namespace std;
typedef pair<int, int> P;
struct edge {
    int to, cost;
};
// other libs
int gcd(int a, int b) {
    if (a % b == 0)
        return b;
    else
        return gcd(b, a % b);
}
int lcm(int a,int b){
    return a*b/gcd(a,b);
}

class dijkstra {
  private:
    int V;
    int d[100002];
    priority_queue<P, vector<P>, greater<>> que;

  public:
    dijkstra(int v, int s, vector<edge> G[]) {
        V = v;
        update(s, G);
    }

    void update(int s, vector<edge> G[]) {
        fill(d, d + V + 1, 100000000000ll);
        d[s] = 0;
        que.push(make_pair(0, s));
        while (!que.empty()) {
            P p = que.top();
            que.pop();
            int v = p.second;
            if (d[v] < p.first)
                continue;
            rep(i, G[v].size()) {
                edge e = G[v][i];
                if (d[e.to] > d[v] + e.cost) {
                    d[e.to] = d[v] + e.cost;
                    que.push(make_pair(d[e.to], e.to));
                }
            }
        }
    }

    int find(int e) {
        return d[e];
    }
};
template<typename T>
class Union_Find {
  private:
    vector<T> t, s;

  public:
    Union_Find(T max_length) {
        rep(i, max_length + 1) {
            t.push_back(i);
            s.push_back(1);
        }
    }

    void Union(T x, T y) {
        if (same(x, y))
            return;
        T tx = Find(x), ty = Find(y);
        if (s[tx] < s[ty]) {
            s[ty] += s[tx];
            t[tx] = ty;
        }
        else if (s[tx] > s[ty]) {
            s[tx] += s[ty];
            t[ty] = tx;
        }
        else if (tx > ty) {
            t[tx] = ty;
            s[ty] += s[tx];
        }
        else {
            t[ty] = tx;
            s[tx] += s[ty];
        }
    }
    T Find(T n) {
        if (t[n] == n)
            return n;
        else
            return t[n] = Find(t[n]);
    }

    bool same(T x, T y) {
        return Find(x) == Find(y);
    }

    T get_Size(size_t a) {
        return s[a];
    }
};
bool ifPrime(int n){
    if(n==1)
        return false;
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0)
            return false;
    }
    return true;
}
vector<int> primeFactorization(int n) {
    vector<int> re;
    int cnt = 2;
    while (n != 1) {
        if (n % cnt != 0) {
            cnt++;
            continue;
        }
        n /= cnt;
        re.push_back(cnt);
    }
    return re;
}
signed main(signed argc, char* argv[]) {
    int n;
    cin>>n;
    vector<int> b(n-1);
    int ans=0;
    rep(i,n-1){
        cin>>b[i];
        if(i){
            ans+=min(b[i-1],b[i]);
        }
        else{
            ans+=b[i];
        }
    }
    ans+=b[n-2];
    cout<<ans<<endl;
}