human
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.
impor
assistant
#Here is Python code to find the total number of continuous subsequences (subsegments) [l, r] (l ≤ r) of the original bracket sequence, which are regular bracket sequences.
t 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()