CF1495B.Let's Go Hiking

On a weekend, Qingshan suggests that she and her friend Daniel go hiking. Unfortunately, they are busy high school students, so they can only go hiking on scratch paper.

A permutation $p$ is written from left to right on the paper. First Qingshan chooses an integer index $x$ ($1\le x\le n$) and tells it to Daniel. After that, Daniel chooses another integer index $y$ ($1\le y\le n$, $y \ne x$).

The game progresses turn by turn and as usual, Qingshan moves first. The rules follow:

• If it is Qingshan's turn, Qingshan must change $x$ to such an index $x'$ that $1\le x'\le n$, $|x'-x|=1$, $x'\ne y$, and $p_{x'} \lt p_x$ at the same time.
• If it is Daniel's turn, Daniel must change $y$ to such an index $y'$ that $1\le y'\le n$, $|y'-y|=1$, $y'\ne x$, and $p_{y'} \gt p_y$ at the same time.

The person who can't make her or his move loses, and the other wins. You, as Qingshan's fan, are asked to calculate the number of possible $x$ to make Qingshan win in the case both players play optimally.

一个周末，青衫建议她和朋友丹尼尔一起去远足。不幸的是，他们都是忙碌的高中生，因此只能在草稿纸上“远足”。

一张排列 $p$ 从左到右写在纸上。首先，青衫选择一个整数下标 $x$（$1\le x\le n$），并将其告诉丹尼尔；随后，丹尼尔选择另一个整数下标 $y$（$1\le y\le n$，且 $y \ne x$）。

游戏轮流进行，照例由青衫先手。规则如下：

• 若轮到青衫行动，她必须将 $x$ 改为某个下标 $x'$，满足：$1\le x'\le n$，$|x'-x|=1$，$x'\ne y$，且同时有 $p_{x'} \lt p_x$；
• 若轮到丹尼尔行动，他必须将 $y$ 改为某个下标 $y'$，满足：$1\le y'\le n$，$|y'-y|=1$，$y'\ne x$，且同时有 $p_{y'} \gt p_y$。

无法进行合法移动的一方判负，另一方获胜。作为青衫的粉丝，你需要计算：在双方均采取最优策略的前提下，有多少个可能的初始 $x$ 值能使青衫获胜。

The first line contains a single integer $n$ ($2\le n\le 10^5$) — the length of the permutation.

The second line contains $n$ distinct integers $p_1,p_2,\dots,p_n$ ($1\le p_i\le n$) — the permutation.

第一行包含一个整数 $n$（$2\le n\le 10^5$）—— 排列的长度。

第二行包含 $n$ 个互不相同的整数 $p_1,p_2,\dots,p_n$（$1\le p_i\le n$）—— 该排列。

Print the number of possible values of $x$ that Qingshan can choose to make her win.

输出青衫能够选择以确保获胜的 $x$ 的可能取值个数。

In the first test case, Qingshan can only choose $x=3$ to win, so the answer is $1$.

In the second test case, if Qingshan will choose $x=4$, Daniel can choose $y=1$. In the first turn (Qingshan's) Qingshan chooses $x'=3$ and changes $x$ to $3$. In the second turn (Daniel's) Daniel chooses $y'=2$ and changes $y$ to $2$. Qingshan can't choose $x'=2$ because $y=2$ at this time. Then Qingshan loses.

在第一个测试用例中，Qingshan 只能选择 $x=3$ 才能获胜，因此答案为 $1$。

在第二个测试用例中，若 Qingshan 选择 $x=4$，Daniel 可以选择 $y=1$。在第一回合（Qingshan 的回合）中，Qingshan 选择 $x'=3$，并将 $x$ 改为 $3$；在第二回合（Daniel 的回合）中，Daniel 选择 $y'=2$，并将 $y$ 改为 $2$。此时由于 $y=2$，Qingshan 无法选择 $x'=2$。于是 Qingshan 失败。