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: '''
      Auther: ghoshashis545 Ashis Ghosh
      College: Jalpaiguri Govt Enggineering College
  
  '''
  from os import path
  from io import BytesIO, IOBase
  import sys
  from heapq import heappush,heappop
  from functools import cmp_to_key as ctk
  from collections import deque,Counter,defaultdict as dd 
  from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
  from itertools import permutations
  from datetime import datetime
  from math import ceil,sqrt,log,gcd
  def ii():return int(input())
  def si():return input().rstrip()
  def mi():return map(int,input().split())
  def li():return list(mi())
  abc='abcdefghijklmnopqrstuvwxyz'
  mod=1000000007
  #mod=998244353
  inf = float("inf")
  vow=['a','e','i','o','u']
  dx,dy=[-1,1,0,0],[0,0,1,-1]
  
  def bo(i):
      return ord(i)-ord('0')
  
  file = 1
  def ceil(a,b):
      return (a+b-1)//b
  
  
  
  
  def solve():
  
      
      # for _ in range(1,ii()+1):
      n = ii()
      a = li()
      ans = 0
      for i in range(0,n,2):
          c = 0
          open_needed = 0 
          for j in range(i+1,n):
              # print(c,open_needed,ans)
              if j%2==0:
                  c += a[j]
              else:
                  rem_op = a[i]-open_needed
                  rem_cls = a[j]-c
                  ans += max(0,min(min(a[j],rem_cls+1),min(a[i],rem_op+1)))
                  c-=a[j]
                  if c<0:
                      open_needed += abs(c)
                      c = 0
          # print()
      print(ans)
  
          
          
  
  
  
  
  
  
  
              
  
  
  
  
  
      
          
  
  
  
  
  
  
  
  
          
  if __name__ =="__main__":
  
      if(file):
      
         if path.exists('input.txt'):
             sys.stdin=open('input.txt', 'r')
             sys.stdout=open('output.txt','w')
         else:
             input=sys.stdin.readline
      solve()
  