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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
#include <stdlib.h>

typedef unsigned long long int uint64;
typedef long long int          int64;
 
#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define ABS(a) ((a)>(0)?(a):-(a))

void get_N (int64 *n)
{
  int cnt;
  
  cnt = scanf("%lld", n);

  return;
}

void get_NK (int64 *n, int64 *k)
{
  int cnt;
  
  cnt = scanf("%lld %lld", n, k);

  return;
}

typedef struct _tmp_t {
  int64    a;
  int64    b;
} tmp_t;


int compare_a(const void *a, const void *b)
{
  tmp_t *x;
  tmp_t *y;

  x = (tmp_t *)a;
  y = (tmp_t *)b;

  if (x->a > y->a) return 1;
  if (x->a < y->a) return -1;

  return 0;
}

int compare_b(const void *a, const void *b)
{
  tmp_t *x;
  tmp_t *y;

  x = (tmp_t *)a;
  y = (tmp_t *)b;

  if (x->b > y->b) return 1;
  if (x->b < y->b) return -1;

  return 0;
}

void qsort_a (tmp_t *array, int size)
{
  qsort(array, size, sizeof(tmp_t), compare_a);
}

void qsort_b (tmp_t *array, int size)
{
  qsort(array, size, sizeof(tmp_t), compare_b);
}

int A[100];
int B[100];

int main()
{
  int cnt;
  int64 N;
  int64 i;
  int64 result = 0;

  get_N(&N);

  for (i=0; i<(N-1); i++){
    cnt = scanf("%d", &B[i]);
    A[i] = -1;
  }
  A[N-1] = -1;

  for (i=0; i<(N-1); i++) {
    if (A[i] == -1) {
      A[i] = B[i];
    }
    if (A[i] > B[i]) {
      A[i] = B[i];
    }
    if (A[i+1] == -1) {
      A[i+1] = B[i];
    }
    if (A[i+1] > B[i]) {
      A[i+1] = B[i];
    }
  }
// A 0 0   10 10 10 23
// B 0 153 10 10 23

  for (i=0; i<N; i++) {
    result += A[i];
  }

  printf("%lld", result);

  return 0;
}