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 itertools import accumulate
  n = int(input())
  
  arr = [int(c) for c in input().split(' ')]
  arr_par = [arr[i] if not i%2 else -arr[i] for i in range(len(arr))]
  acc = list(accumulate(arr_par))
  
  ctr = 0
  
  def d_inc(d, i):
      if i in d:
          d[i] += 1
      else:
          d[i] = 1
  
  def summ(i):
      return i*(i+1)/2
  def corr(i):
      return summ(i+1) - (i+1)
  
  def handle(d, ths):
      ad = 0
      for k in d:
          if ths < k:
              #print('closing', k)
              if d[k]:
                  #print('CORR', corr(d[k]))
                  ad += corr(d[k])
              d[k] = 0
      return ad
  
  
  open_reg = 0
  mn = 0
  ths = 0
  special = {}
  #print(arr)
  for i in range(n):
      ths += arr_par[i]
      #print(ths, arr[i])
      if not i%2:
          open_reg += arr[i]
          #print('upwards', open_reg)
      else:
          if arr[i] <= open_reg:
              ctr += arr[i]
              #print('added:', arr[i])
              open_reg -= arr[i]
          else:
              ctr += open_reg
              open_reg = 0
              #print('added:' ,open_reg)
  
          corr_val = handle(special, ths)
          ctr += corr_val
          #print('corr:' ,corr_val)
          #print(' ')
          if ths < mn:
              mn = ths
          else:
              d_inc(special, ths)
  
  #print('final closing')
  ad = 0
  d = special
  for k in d:
      #print('closing', k)
      if d[k]:
          #print('CORR', corr(d[k]))
          ad += corr(d[k]-1)
      d[k] = 0
  
  ctr += ad
  
  print(int(ctr))
  
  
  