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: import itertools as _itertools
  import math as _math
  import sys as _sys
  import time as _time
  
  
  def f(c_seq):
      c_seq = [+c if i % 2 == 0 else -c for i, c in enumerate(c_seq)]
      prefix_sums = tuple(_itertools.accumulate(c_seq))
      segs_minimums = _precompute_minimums(prefix_sums)
      
      result = 0
      for l in range(0, len(c_seq), 2):
          for r in range(l + 1, len(c_seq), 2):
              opening_n = abs(c_seq[l])
              closing_n = abs(c_seq[r])
              
              if r > l + 1:
                  min_sum_between = segs_minimums[l+1, r-1] - prefix_sums[l]
                  max_sum_between = prefix_sums[r-1] - segs_minimums[l, r-2]
              else:
                  min_sum_between = 0
                  max_sum_between = 0
              
              if max_sum_between > 0:
                  closing_n -= abs(max_sum_between) - 1
              if min_sum_between < 0:
                  opening_n -= abs(min_sum_between) - 1
              
              curr_variants_n = min(opening_n, closing_n)
              if curr_variants_n >= 0:
                  result += curr_variants_n
      
      return result
  
  
  def _precompute_minimums(sequence):
      result = dict()
      for l in range(len(sequence)):
          curr_minimum = _math.inf
          for r in range(l, len(sequence)):
              curr_minimum = min(curr_minimum, sequence[r])
              result[l, r] = curr_minimum
      return result
  
  
  def _main():
      start_time = _time.time()
      [c_n] = _read_ints()
      c_seq = tuple(_read_ints())
      assert len(c_seq) == c_n
      result = f(c_seq)
      print(result)
      time_limit = 1
      while _time.time() - start_time < time_limit - 1/0x10:
          _time.sleep(1/0x100)
  
  
  def _read_ints(file=_sys.stdin):
      return map(int, file.readline().split())
  
  
  if __name__ == '__main__':
      _main()
  