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: from sys import stdout, stdin 
  input = stdin.readline
  
  def output(answer):
      stdout.write("{}\n".format(answer))
  
  def transfer(stack,other,flag):
      if flag == 0:
          while len(other) > 0:
              stack.append(other.pop())
      else:
          while len(other) > 0:
              v = other.pop()
              if v > 0:
                  stack.append(v)
  
  
  
  for _ in range(1):
  
      n = int(input())
      arr = list(map(int,input().split()))
  
      stack = [None]
      count = 0
  
      for index in range(n):
          # print(stack,count,arr[index])
          if index%2 == 0:
              stack.append(arr[index])
              continue
          curr = arr[index]
          other = []
          flag = 0
          while stack[-1] != None and curr != 0 and stack[-1] <= curr:
  
              value = stack.pop()
              # print(value,curr,index)
              if value < 0:
                  count += 1
                  other.append(value)
              elif value > curr:
                  count += curr
                  # print(count,curr,index)
                  stack.append(value - curr)
                  stack.append(-1)
                  curr = 0
                  break
              elif value <= curr:
                  if value == curr:
                      count += curr
                      curr = 0
                      itr = len(stack) - 1
                      while itr > 0 and stack[itr] != None and stack[itr] < 0:
                          count += 1
                          itr -= 1
                      stack.append(-1)
                  else:
                      count += value 
                      other.append(-1)
                      curr -= value
                      flag = 1
  
  
          if stack[-1] == None:
              transfer(stack, other, flag)
              if curr > 0:
                  stack.append(None)
          elif curr == 0:
              transfer(stack, other, flag)
          elif curr != 0:
              value = stack.pop()
              stack.append(value - curr)
              stack.append(-1)
              count += curr
              transfer(stack, other, 1)
  
      output(count)