An $O(n)$ `Partition` algorithm partitions an array
`A` around a `pivot` element (`pivot`
is a member of `A`) into three parts: a
left sub-array that contains elements that are $\leq $ `pivot`, the `pivot`
itself, and a right sub-array that contains elements that are
$>$ `pivot`.

A `Partition` algorithm is an
integral part of a popular sorting algorithm `Quicksort`. Usually, the choice of `pivot` is randomized so that `Quicksort` has an expected $O(n \log n)$ running time
performance.

Now the actual job in this problem is this: Starting from an
array `A` that has $n$ distinct integers, we partition
`A` using one of the member of
`A` as pivot to produce a transformed
array `A’`. Given this transformed
array `A’`, your job is to count how
many integers that could have been the chosen pivot.

For example, if `A’` = $\{ 2, 1, 3, 4, 7, 5, 6, 8\} $, then 3
integers: $\{ 3, 4, 8\} $
could have been the pivot of partition, e.g. $\{ 2, 1, 3\} $ to the left of integer
$4$ are smaller than
$4$ and $\{ 7, 5, 6, 8\} $ to the right of
integer $4$ are greater
than $4$. However, the
other 5 integers cannot possibly be the pivot, e.g. integer
$7$ cannot possibly be the
pivot as there are $\{ 5, 6\}
$ to the right of integer $7$.

The input consists of two lines. The first line contains integer $n$ ($3 \leq n \leq 100\, 000$). The second line contains $n$ distinct 32-bit signed integers that describes the transformed array $A’$.

Output the required answer as a single integer in one line.

Sample Input 1 | Sample Output 1 |
---|---|

8 2 1 3 4 7 5 6 8 |
3 |

Sample Input 2 | Sample Output 2 |
---|---|

7 1 2 3 4 5 7 6 |
5 |