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 sys
  
  input = sys.stdin.readline
  
  
  def ri(): return [int(i) for i in input().split()]
  
  
  def common(amin, amax, bmin, bmax):
      if amax < bmin or bmax < amin:
          return 0
      return min(amax, bmax) - max(amin, bmin) + 1
  
  
  def solve(Aseg, Bseg, fminopen, faddsopen):
      # from A must take at least fminopen open-brackets
      # from B must close at least faddsopen close-brackets
      if fminopen > Aseg:
          return 0
      amin = max(1, fminopen) + faddsopen
      amax = Aseg + faddsopen
      return common(amin, amax, 1, Bseg)
  
  
  def main():
      t = 1
  
      for _ in range(t):
          n = ri()[0]
          a = ri()
          res = 0
          for st_seg_idx in range(0, n, 2):
              f_min_open = 0
              f_adds_open = 0
              seg_result = 0
              for end_seg_idx in range(st_seg_idx + 1, n):
                  if end_seg_idx % 2 == 0:
                      # ( segment
                      f_adds_open += a[end_seg_idx]
                  else:
                      # ) segment
                      # try to obtain closed () here
                      seg_result += solve(a[st_seg_idx], a[end_seg_idx], f_min_open, f_adds_open)
                      # add this segment to stack
                      f_adds_open -= a[end_seg_idx]
                      if f_adds_open < 0:
                          f_min_open = max(f_min_open, -f_adds_open)
  
                  if f_min_open > a[st_seg_idx]:
                      break
              # print(f'seg result for [{a[st_seg_idx]}]: {seg_result}')
              res += seg_result
          print(res)
  
  
  main()
  