description: <image>
  
  William has a favorite bracket sequence. Since his favorite sequence is quite big he provided it to you as a sequence of positive integers c_1, c_2, ..., c_n where c_i is the number of consecutive brackets "(" if i is an odd number or the number of consecutive brackets ")" if i is an even number.
  
  For example for a bracket sequence "((())()))" a corresponding sequence of numbers is [3, 2, 1, 3].
  
  You need to find the total number of continuous subsequences (subsegments) [l, r] (l ≤ r) of the original bracket sequence, which are regular bracket sequences.
  
  A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, sequences "(())()", "()" and "(()(()))" are regular, while ")(", "(()" and "(()))(" are not.
  
  Input
  
  The first line contains a single integer n (1 ≤ n ≤ 1000), the size of the compressed sequence.
  
  The second line contains a sequence of integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9), the compressed sequence.
  
  Output
  
  Output a single integer — the total number of subsegments of the original bracket sequence, which are regular bracket sequences.
  
  It can be proved that the answer fits in the signed 64-bit integer data type.
  
  Examples
  
  Input
  
  
  5
  4 1 2 3 1
  
  
  Output
  
  
  5
  
  
  Input
  
  
  6
  1 3 2 1 2 4
  
  
  Output
  
  
  6
  
  
  Input
  
  
  6
  1 1 1 1 2 2
  
  
  Output
  
  
  7
  
  Note
  
  In the first example a sequence (((()(()))( is described. This bracket sequence contains 5 subsegments which form regular bracket sequences:
  
    1. Subsequence from the 3rd to 10th character: (()(()))
    2. Subsequence from the 4th to 5th character: ()
    3. Subsequence from the 4th to 9th character: ()(())
    4. Subsequence from the 6th to 9th character: (())
    5. Subsequence from the 7th to 8th character: ()
  
  
  
  In the second example a sequence ()))(()(()))) is described.
  
  In the third example a sequence ()()(()) is described.
solution: INF = 10**18
  import sys
  import pprint
  
  # sys.setrecursionlimit(10 ** 9)
  
  def input(): return sys.stdin.readline().rstrip("\r\n")
  
  def maps(): return [int(i) for i in input().split()]
  
  # lOOKOUT FOR THE EDGE CASES
  
  
  import logging
  from logging import getLogger
  
  logging.basicConfig(
      format="%(message)s",
      level=logging.WARNING,
  )
  logger = getLogger(__name__)
  logger.setLevel(logging.INFO)
  
  
  def debug(msg, *args):
      logger.info(f'{msg}={pprint.pformat(args)}')
  
  
  # logger.info(f"dp_list=\n{pprint.pformat(dp_list)}")
  def solve1():
      n, = maps()
      a = [*maps()]
      ans = 0
      for i in range(0, n, 2):
          bal, minbal = 0, 0
          for j in range(i + 1, n):
              if j % 2:
                  left = -minbal
                  right = bal - minbal  # parens needed to make the expression correct
                  left = max(1, left)
                  right = max(1, right)  # balance it if it's the first 2 parens
                  if left <= a[i] and right <= a[j]:
                      # cut the balance out and check how many can be formed
                      ans += min(a[i] - left, a[j] - right) + 1
  
              if j % 2 == 0:
                  bal += a[j]
              else:
                  bal -= a[j]
  
              minbal = min(minbal, bal)
      print(ans)
  
  
  def solve2():
      n, = maps()
      a = [*maps()]
      ans = 0
      st = []
      for i in range(n):
          if i % 2 == 0:
              if not st or st[-1][0] != '(':
                  st.append(['(', a[i]])
              else:
                  st[-1][1] += a[i]
  
          else:
              while st:
                  if st and st[-1][0] == '(':
                      if st[-1][1] <= a[i]:
                          a[i] -= st[-1][1]
                          st[-1][0] = '()'
                      else:
                          st[-1][1] -= a[i]
                          st.append(['()', a[i]])
                          a[i] = 0
  
                  if a[i] == 0:
                      # debug("stack when a[i] == 0", st)
                      # it breaks because it's still a vaild expression and more parens can be
                      # added to it , to make a bigger parens
                      break
                  # here it doesn't break because no more opening parens can be found and thus balance is <=0
                  acc = 0
  
                  while st and st[-1][0] == '()':
                      ans += st[-1][1] + acc
                      acc += 1
                      # debug("stack", st)
                      st.pop()
                      # debug("stack", st)
      # debug("", st)
      acc = 0
      while st:
          if st[-1][0] == '()':
              ans += st[-1][1] + acc
              acc += 1
          else:
              acc = 0
              # it doesn't accumulate because it's not a consecutive correct expressions but in blocks each independent
          st.pop()
  
      print(ans)
  
  
  solve2()
  