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: Wrong Answer)

#include <iostream>
#include<algorithm>
#include<string>
#include <cmath>
#include <vector>
#include <map>
#include <cstdio>
#include <iomanip>
#include<set>
#include <numeric>
#pragma region Macros
#define int long long
#define double long double
constexpr int MOD = 1000000007;
constexpr double PI = 3.14159265358979323846;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
#define LAST(x) x[x.size()-1]
#define ALL(x) (x).begin(),(x).end()
#define swap(a,b) (a += b,b = a - b,a -= b)
#define DIVCEIL(a,b) ((a+b-1)/b)
//int CHMAX

int FACT(int a) {
	if (a == 0)
		return 1;
	else
		return a * FACT(a - 1);
}
int MODFACT(int a, int mod) {
	if (a == 0)
		return 1;
	else
		return (a * FACT(a - 1))%mod;
}
int nPr(int n, int r) {
	int s = n - r + 1;
	int sum = 1;
	for (int i = s; i <= n; i++)
		sum *= i;
	return sum;
}
int MODnPr(int n, int r, int mod) {
	int s = n - r + 1;
	int sum = 1;
	for (int i = s; i <= n; i++) {
		sum *= i;
		sum = sum % MOD;
	}
	return sum;
}
//int nCr(int n, int r)
int nCr2(int n, int r) {
	return FACT(n) / (FACT(r) * FACT(n - r));
}
int GCD(int a, int b)
{
	if (a < b)
		swap(a, b);
	if (b == 0)
		return a;
	if (a % b == 0)
		return b;
	return GCD(b, a % b);
}
int LCM(int a, int b)
{
	return  a * b / GCD(a, b);
}
int NUMOFDIV(int n) {
	int ans = 0;
	REP(i, n) {
		if (n % i == 0)
			ans++;
	}
	return ans;
}
int CEIL1(int n) {
	return (n + 9) / 10 * 10;
}
int DIGITSUM(int n) {
	int sum = 0, dig;
	while (n) {
		dig = n % 10;
		sum += dig;
		n /= 10;
	}
	return sum;
}
int DIVTIME(int n, int k) {
	int div = 0;
	while (n % k == 0) {
		div++;
		n /= k;
	}
	return div;
}
int MODPOW(int a, int n, int mod) {
	// a^n mod
	int ans = 1;
	while (n > 0) {
		if (n & 1)
			ans = ans * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return ans;
}
double LOG(int a, int b) {
	return log(b) / log(a);
}
inline bool BETWEEN(int x, int min, int max) {
	if (min <= x && x <= max)
		return true;
	else
		return false;
}
inline bool PRIMEOR(int x) {
	if (x == 1)
		return false;
	if (x == 2)
		return true;
	if (x % 2 == 0)
		return false;
	double sqrtx = sqrt(x);
	for (int i = 3; i <= sqrtx; i += 2) {
		if (x % i == 0)
			return false;
	}
	return true;
}
bool SQRTOR(int n) {
	if (sqrt(n) == (int)sqrt(n))
		return true;
	else
		return false;
}
// 順位付け
using namespace std;
#pragma endregion
signed main() {
	int N, ans = 0;
	cin >> N;
	vector<int>a(N);
	rep(i, N)
		cin >> a[i];
	ans += a[0];
	ans += a[N];
	rep(i, N - 1)
		ans += min(a[i], a[i + 1]);
	cout << ans;
}