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: """
                              pppppppppppppppppppp
                           ppppp  ppppppppppppppppppp
                        ppppppp    ppppppppppppppppppppp
                        pppppppp  pppppppppppppppppppppp
                        pppppppppppppppppppppppppppppppp
                                pppppppppppppppppppppppp
         ppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppp
        pppppppppppppppppppppppppppppppppppppppppppppppp  ppppppppppppppppppppp
       ppppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppppp
      ppppppppppppppppppppppppppppppppppppppppppppppp    pppppppppppppppppppppppp
     pppppppppppppppppppppppppppppppppppppppppppppp     pppppppppppppppppppppppppp
    ppppppppppppppppppppppppppppppppppppppppppppp      pppppppppppppppppppppppppppp
    pppppppppppppppppppppppppppppppp               pppppppppppppppppppppppppppppppp
    pppppppppppppppppppppppppppp     pppppppppppppppppppppppppppppppppppppppppppppp
    ppppppppppppppppppppppppppp    pppppppppppppppppppppppppppppppppppppppppppppppp
      pppppppppppppppppppppppp  pppppppppppppppppppppppppppppppppppppppppppppppppp
       ppppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppppp
        pppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppp
         ppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppp
                                pppppppppppppppppppppppp
                                pppppppppppppppppppppppppppppppp
                                pppppppppppppppppppppp  pppppppp
                                ppppppppppppppppppppp    ppppppp
                                   ppppppppppppppppppp  ppppp
                                      pppppppppppppppppppp
  """
  
  
  import sys
  from functools import lru_cache, cmp_to_key
  from heapq import merge, heapify, heappop, heappush, nsmallest
  from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
  from collections import defaultdict as dd, deque, Counter as C
  from itertools import combinations as comb, permutations as perm
  from bisect import bisect_left as bl, bisect_right as br, bisect
  from time import perf_counter
  from fractions import Fraction
  from decimal import Decimal
  # sys.setrecursionlimit(pow(10, 6))
  # sys.stdin = open("input.txt", "r")
  # sys.stdout = open("output.txt", "w")
  
  mod = pow(10, 9) + 7
  # mod2 = 998244353
  def data(): return sys.stdin.readline().strip()
  def out(var): sys.stdout.write(str(var)+"\n")
  def outa(*var, end="\n"): sys.stdout.write(' '.join(map(str, var)) + end)
  def l(): return list(sp())
  def sl(): return list(ssp())
  def sp(): return map(int, data().split())
  def ssp(): return map(str, data().split())
  def l1d(n, val=0): return [val for i in range(n)]
  def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
  
  
  n = int(data())
  arr = l()
  cnt, answer, temp = [[0, 0]], 0, 0
  for i in range(0, n - 1, 2):
      temp += arr[i] - arr[i + 1]
      answer += min(arr[i + 1], temp + arr[i + 1] - cnt[0][0])
      while cnt and cnt[-1][0] > temp:
          answer += cnt[-1][1]
          cnt.pop()
      if cnt and cnt[-1][0] == temp:
          answer += cnt[-1][1]
          cnt[-1][1] += 1
      else:
          if not cnt:
              cnt.append([temp, 0])
          else:
              cnt.append([temp, 1])
  out(answer)
  