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
  ints = (int(x) for x in sys.stdin.read().split())
  sys.setrecursionlimit(3000)
  
   
  class Stack(list):
      ans = 0
      lzeros = 0
      zeros = []
  
      def save(self, n, k):
          #print(n, k)
          self.ans += n + ((k*(k-1))//2)
  
      def read(self, seq):
          it = iter(seq)
          for a,b in zip(it, it):
              self.append(a)
              self.zeros.append(0)
              self.close(b)
          while self:
              self.pop()
              self.save(0, self.zeros.pop())
          self.save(0, self.lzeros)
          self.lzeros = 0
          return self.ans
   
      def close(self, n):
          assert n > 0
          while self and n >= self[-1]:
              n -= self[-1]
              self.save(self.pop(), self.zeros.pop())
              if self:
                  self.zeros[-1] += 1
              else:
                  self.lzeros += 1
          if n > 0:
              if not self:
                  self.save(0, self.lzeros)
                  self.lzeros = 0
              elif n < self[-1]:
                  self.save(n, self.zeros[-1])
                  self[-1] -= n
                  self.zeros[-1] = 1
          return
  
  def main():
      n = next(ints)
      stack = Stack()
      stack.read([next(ints) for i in range(n)])
      print(stack.ans)
      return
  
  
  main()
  