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 (Clang 3.8.0) (judge verdict: Accepted)

#include<iostream>
#include<algorithm>
#include<set>
#include<math.h>
#include<vector>
#include<sstream>
#include<queue>
#include<functional>
#include<bitset>
#include<cstdio>
#include<iomanip>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include <string.h>

using ll = long long;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define reps(i, x) for(int i=1;i<=(int)(x);i++)
#define rrep(i, x) for(int i=((int)(x)-1);i>=0;i--)
#define rreps(i, x) for(int i=(int)(x);i>0;i--)
#define all(x) (x).begin(),(x).end()
#define m0(x) memset(x,0,sizeof(x))
#define vll vector<ll>
#define vi vector<int>
#define vpll vector<pair<ll,ll>>
#define vpi vector<pair<int,int>>
#define mod 1000000007
#define put2darr(arr, row, col) rep(i, row) { rep(j, col) { printf("%18d", arr[i][j]); } cout << endl; }

using namespace std;

template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return 1; } return 0; }

/* 1..10
 * rep:   0 1 2 3 4 5 6 7 8 9
 * reps:  1 2 3 4 5 6 7 8 9 10
 * rrep:  9 8 7 6 5 4 3 2 1 0
 * rreps: 10 9 8 7 6 5 4 3 2 1
 */
int b[100];
int main(){
    int n;
    cin >> n;
    rep(i, n-1){
      cin >> b[i];
    }
    b[n-1] = 1 << 30;
    int ans = b[0];
    rep(i, n -1){
        ans += min(b[i], b[i+1]);
    }
    cout << ans << endl;
}