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: #from itertools import product, permutations, combinations, accumulate, compress
  #from collections import Counter, defaultdict, deque # deque([iterable[, maxlen]]) #appendleft popleft rotate
  #from heapq import heapify, heappop, heappush # func(heapifiedlist, item)
  #from bisect import bisect_left, bisect_right, insort # func(sortedlist, item)
  # TODO: more coping mechanisms...?
  #from sys import setrecursionlimit
  
  import os, sys
  input = lambda: sys.stdin.buffer.readline().rstrip(b'\r\n')
  
  def dbp(*args, **kwargs): # calling with dbp(locals()) is perfectly cromulent
  	print(*args, file=sys.stderr, **kwargs)
  
  a_str = lambda: input().decode('ascii')
  get_int_list = lambda: [int(x) for x in input().split()]
  join_str = lambda l, ch=' ': ch.join((str(x) for x in l))  # ch is a default, not kwarg
  
  def do_thing():
  	n = int(input())
  	alist = get_int_list()
  
  	def slow_solve():
  		s = []
  		for idx, a in enumerate(alist):
  			if idx%2:
  				s.extend([')']*a)
  			else:
  				s.extend(['(']*a)
  		dbp(s)
  		count = 0
  		ls = len(s)
  		for w in range(2, ls+1, 2):
  			for i in range(ls-w+1):			
  				h = 0
  				ok = True
  				sub = s[i:i+w]
  				for c in sub:
  					h += 1 if c == '(' else -1
  					if h < 0:
  						ok = False
  						break
  				if ok and h == 0:
  					count += 1
  					dbp(i, i+w-1, ''.join(sub), count) # 0-idx, r-inc
  		return count
  
  	def regret_solve():
  		basecount = strkcount = 0
  		lefts = 0
  		hstk = [[0, 0]]
  		#dbp('pre:', locals())
  		for i in range(0, n-1, 2):
  			l, r = alist[i:i+2]
  			lefts += l
  			basecount += min(lefts, r)
  			lefts -= r
  			while hstk and hstk[-1][0] > lefts:
  				h, strk = hstk.pop()
  				strkcount += strk
  			if hstk and hstk[-1][0] == lefts:
  				strkcount += hstk[-1][1]
  				hstk[-1][1] += 1
  			else:
  				if hstk:
  					#dbp('sus')
  					hstk.append([lefts, 1])
  
  			if lefts < 0:
  				lefts = 0
  			if not hstk:
  				hstk = [[lefts, 0]]
  			#dbp('post:', locals())
  		#dbp('base:', basecount, 'strkcount:', strkcount)
  		return basecount+strkcount
  
  
  	#sr = slow_solve()
  	r = regret_solve()
  	#dbp('results:', sr, r)
  	#assert(sr == r)
  
  	return r
  
  if __name__ == "__main__":
  	maxcc = 1
  	#maxcc = int(input())
  	for cc in range(maxcc):
  		print(do_thing())
  