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>
using namespace std;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define REP(i, n) for(int i = 1; i < (int)(n); i++)
typedef long long ll;

int dp[100];

//最大公約数
int gcd(int x, int y){
  if(x % y == 0) return 1;
  return gcd(y, x % y);
}
//最小公倍数
int lcm(int x, int y){
  return x * y / gcd(x, y);
}
//フィボナッチ数列
//配列dpを-1で初期化する
int fib(int n){
  if(n == 0 || n == 1) return dp[n] = 1;
  if(dp[n] != -1) return dp[n];
  return dp[n] = fib(n - 1) + fib(n - 2);
}
//階乗計算(factorial)
int fact(int n){
  if(n = 1) return 1;
  return n * fact(n - 1);
}
//累乗計算(Exponentiation)
//x**y
ll expo(int x, int y){
  ll ans = 1;
  rep(i, y){
    ans *= x;
  }
  return ans;
}
//桁数
int keta(ll x){
  if(x == 0) return 1;
  int res = 0;
  while(x){
    x /= 10;
    res++;
  }
  return res;
}


int main(){
  int n, b[102], a[102];
  cin >> n;
  rep(i, n-1) cin >> b[i];

  int ans = 0;

  a[0] = b[0];
  REP(i, n){
    if(i == n-1){
      a[n-1] = b[n-2];
      break;
    }
    a[i] = min(b[i], b[i-1]);
  }
  rep(i, n) {
    ans += a[i];
  }

  cout << ans << endl;
}


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