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 <iostream>
#include <iomanip>
#include <math.h>
#include <string>
#include <algorithm>
#include <ctype.h>
#include <vector>
#include <numeric>
#include <map>
#include <queue>
using namespace std;

#define ll long long int
#define rep(i, a, n) for (int i = a; i < n; i++)
#define INF 1e9
#define LINF 1e17
#define MOD (int)(1e9 + 7)
#define pi 3.141592653589
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vll vector<ll>
#define ALL(a) (a).begin(), (a).end()
#define sort_v(a) sort(a.begin(), a.end())
#define fi first
#define se second

void print(bool c)
{
    if (c)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}

void Print(bool c)
{
    if (c)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
}

ll gcdll(ll a, ll b)
{
    while (1)
    {
        if (a < b)
            swap(a, b);
        if (b == 0)
            break;
        a %= b;
    }
    return a;
}

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

bool cmp(const pii &a, const pii &b)
{
    if (a.second != b.second)
        return a.second < b.second;
    return a.first < b.first;
}

int main()
{
    int n;
    cin >> n;
    vi b(n - 1);
    rep(i, 0, n - 1) cin >> b[i];
    vi a;
    a.push_back(b[0]);
    rep(i, 0, n - 2)
    {
        a.push_back(min(b[i], b[i + 1]));
    }
    a.push_back(b[n - 2]);
    int ans = 0;
    rep(i, 0, n)
    {
        ans += a[i];
    }
    cout << ans << endl;
}