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.
Solve the task in PYTHON3.

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What is a Python program to find the total number of continuous subsequences (subsegments) [l, r] (l ≤ r) of the original bracket sequence, which are regular bracket sequences?
assistant
#Here is Python program to find the total number of continuous subsequences (subsegments) [l, r] (l ≤ r) of the original bracket sequence, which are regular bracket sequences.

import sys

input = sys.stdin.readline

# def sv1(ls, start):
#     sample = []
#     for _ in ls:
#         sample.append(start)
#         start = 1 - start
#     memo = []
#     for i in range(len(ls)):
#         if ls[i] != sample[i]:
#             memo.append(i)
#     if not memo:
#         return 0
#     if len(memo) % 2 != 0:
#         raise RuntimeError
#     ret = 0
#     for i in range(0, len(memo), 2):
#         ret += abs(memo[i] - memo[i + 1])
#     return ret
#
#
# def sv(ls):
#     ls = [1 if i % 2 == 1 else 0 for i in ls]
#     odd = sum(ls)
#     even = len(ls) - odd
#     if abs(odd - even) > 1:
#         return -1
#     if odd > even:
#         return sv1(ls, 1)
#     elif odd < even:
#         return sv1(ls, 0)
#     else:
#         return min(sv1(ls, 0), sv1(ls, 1))

# def main():
#     for _ in range(int(input())):
#         input()
#         ls = list(map(int, input().split()))
#         print(sv(ls))

def sv(ls):
    ret = 0
    deep = 0
    stack = []
    for i in range(len(ls)):
        v = ls[i]
        if i % 2 == 0:
            deep += v
        else:
            append = True
            if v > deep:
                append = False
            v = min(v, deep)
            ret += v
            deep -= v
            memo = None
            while stack:
                depth, vv = stack[-1]
                if depth < deep:
                    break
                depth, vv = stack.pop()
                if append and depth == deep:
                    memo = [depth, vv + 1]
                ret += vv
            if append:
                if memo:
                    stack.append(memo)
                else:
                    stack.append([deep, 1])
    return ret

def main():
    input()
    a = list(map(int, input().split()))
    print(sv(a))

main()