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
  from bisect import *
  from collections import deque
  #from functools import *
  #from fractions import Fraction as f
  from copy import *	
  from heapq import *
  #from math import *
  from itertools import permutations as prm,product
  from random import *
  def eprint(*args):
      print(*args, file=sys.stderr)
  zz=1
   
  sys.setrecursionlimit(10**5)
  if zz:
  	input=sys.stdin.readline
  else:	
  	sys.stdin=open('input.txt', 'r')
  	sys.stdout=open('all.txt','w')
  di=[[-1,0],[1,0],[0,1],[0,-1]]
  
  def fori(n):
  	return [fi() for i in range(n)]	
  def inc(d,c,x=1):
  	d[c]=d[c]+x if c in d else x
  def ii():
  	return input().rstrip()	
  def li():
  	return [int(xx) for xx in input().split()]
  def fli():
  	return [float(x) for x in input().split()]	
  def dadd(d,p,val):
  	if p in d:
  		d[p].append(val)
  	else:
  		d[p]=[val]		
  def gi():	
  	return [xx for xx in input().split()]
  def gtc(tc,*ans):
  	print("Case #"+str(tc)+":",*ans)	
  def cil(n,m):
  	return n//m+int(n%m>0)	
  def fi():
  	return int(input())
  def pro(a): 
  	return reduce(lambda a,b:a*b,a)		
  def swap(a,i,j): 
  	a[i],a[j]=a[j],a[i]	
  def bits(i,n):
  	p=bin(i)[2:]
  	return (n-len(p))*"0"+p	
  def prec(a,pre):
  	for i in a:
  		pre.append(pre[-1]+i)
  	pre.pop(0)	
  def YN(flag):
  	print("YES" if flag else "NO")	
  def si():
  	return list(input().rstrip())	
  def mi():
  	return 	map(int,input().split())			
  def gh():
  	sys.stdout.flush()
  def isvalid(i,j,n,m):
  	return 0<=i<n and 0<=j<m 
  def bo(i):
  	return ord(i)-ord('a')	
  def graph(n,m):
  	for i in range(m):
  		x,y=mi()
  		a[x].append(y)
  		a[y].append(x)
  
  
  
  t=1
  INF=10**18
  uu=t
  mod=10**9+7
  
  		
  			
  while t>0:
  	t-=1
  	n=fi()
  	a=li()
  	ans=0
  	for i in range(0,n,2):
  		b=0
  		minb=10**18
  		for j in range(i,n):
  			if j%2==0:
  				b+=a[j]
  			else:
  				#sb,minb,curb,a[i]
  				ans+=max(0,-max(0,(b-a[j]))+min(a[i]-1,minb)+1)
  				b-=a[j]
  				if b<0:
  					break	
  			minb=min(minb,b)
  			#print(i,j,b,minb,ans)
  	print(ans)	