Alice got an array of length $n$ as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen $m$ changes of the following form. For some integer numbers $x$ and $d$, he chooses an arbitrary position $i$ ($1 \le i \le n$) and for every $j \in [1, n]$ adds $x + d \cdot dist(i, j)$ to the value of the $j$\-th cell. $dist(i, j)$ is the distance between positions $i$ and $j$ (i.e. $dist(i, j) = |i - j|$, where $|x|$ is an absolute value of $x$).
For example, if Alice currently has an array $[2, 1, 2, 2]$ and Bob chooses position $3$ for $x = -1$ and $d = 2$ then the array will become $[2 - 1 + 2 \cdot 2,~1 - 1 + 2 \cdot 1,~2 - 1 + 2 \cdot 0,~2 - 1 + 2 \cdot 1]$ = $[5, 2, 1, 3]$. Note that Bob can't choose position $i$ outside of the array (that is, smaller than $1$ or greater than $n$).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
## Input
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^5$) — the number of elements of the array and the number of changes.
Each of the next $m$ lines contains two integers $x_i$ and $d_i$ ($-10^3 \le x_i, d_i \le 10^3$) — the parameters for the $i$\-th change.
## Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed $10^{-6}$.
[samples]
Alice 再次收到了一个长度为 $n$ 的数组作为生日礼物！这已经是连续第三年了！
更令人失望的是，这个数组极其无聊，全部由零组成。Bob 决定对数组进行一些修改，以让 Alice 开心起来。
Bob 选择了 $m$ 次如下形式的修改。对于某些整数 $x$ 和 $d$，他选择一个任意位置 $i$（$1 lt.eq i lt.eq n$），并对每个 $j in [ 1, n ]$，将 $x + d dot.op d i s t (i, j)$ 加到第 $j$ 个单元格的值上。$d i s t (i, j)$ 是位置 $i$ 和 $j$ 之间的距离（即 $d i s t (i, j) = | i -j |$，其中 $| x |$ 表示 $x$ 的绝对值）。
例如，如果 Alice 当前的数组是 $[ 2, 1, 2, 2 ]$，而 Bob 选择位置 $3$，并令 $x = -1$、$d = 2$，则数组将变为 $[ 2 -1 + 2 dot.op 2, med 1 -1 + 2 dot.op 1, med 2 -1 + 2 dot.op 0, med 2 -1 + 2 dot.op 1 ]$ = $[ 5, 2, 1, 3 ]$。注意，Bob 不能选择超出数组范围的位置（即小于 $1$ 或大于 $n$）。
当数组中的元素尽可能大时，Alice 会最快乐。Bob 认为元素的算术平均值是一个合适的度量标准。
Bob 能达到的最大算术平均值是多少？
第一行包含两个整数 $n$ 和 $m$（$1 lt.eq n, m lt.eq 10^5$）——数组的元素个数和修改次数。
接下来的 $m$ 行每行包含两个整数 $x_i$ 和 $d_i$（$-10^3 lt.eq x_i, d_i lt.eq 10^3$）——第 $i$ 次修改的参数。
请输出 Bob 能达到的最大算术平均值。
你的答案若绝对误差或相对误差不超过 $10^(-6)$，则被视为正确。
## Input
第一行包含两个整数 $n$ 和 $m$（$1 lt.eq n, m lt.eq 10^5$）——数组的元素个数和修改次数。接下来的 $m$ 行每行包含两个整数 $x_i$ 和 $d_i$（$-10^3 lt.eq x_i, d_i lt.eq 10^3$）——第 $i$ 次修改的参数。
## Output
请输出 Bob 能达到的最大算术平均值。你的答案若绝对误差或相对误差不超过 $10^(-6)$，则被视为正确。
[samples]
**Definitions**
Let $ n, m \in \mathbb{Z}^+ $ denote the length of the array and the number of changes, respectively.
Let $ A = (a_1, a_2, \dots, a_n) $ be the array, initially $ a_j = 0 $ for all $ j \in \{1, \dots, n\} $.
For each change $ i \in \{1, \dots, m\} $, Bob selects a position $ p_i \in \{1, \dots, n\} $ and adds to each element $ a_j $ the value:
$$
x_i + d_i \cdot |p_i - j|
$$
**Constraints**
1. $ 1 \le n, m \le 10^5 $
2. $ -10^3 \le x_i, d_i \le 10^3 $ for all $ i \in \{1, \dots, m\} $
3. For each change $ i $, the position $ p_i $ is chosen optimally from $ \{1, \dots, n\} $ to maximize the final average.
**Objective**
Maximize the arithmetic mean of the final array:
$$
\max_{p_1, \dots, p_m \in \{1,\dots,n\}} \left( \frac{1}{n} \sum_{j=1}^n \sum_{i=1}^m \left( x_i + d_i \cdot |p_i - j| \right) \right)
$$
Equivalently, since the mean is linear:
$$
\max_{p_1, \dots, p_m \in \{1,\dots,n\}} \left( \sum_{i=1}^m x_i + \frac{1}{n} \sum_{i=1}^m d_i \sum_{j=1}^n |p_i - j| \right)
$$
For each change $ i $, define:
$$
f_i(p) = \sum_{j=1}^n |p - j|, \quad p \in \{1, \dots, n\}
$$
Then the objective becomes:
$$
\sum_{i=1}^m x_i + \sum_{i=1}^m d_i \cdot \max_{p \in \{1,\dots,n\}} f_i(p)
$$
(Note: if $ d_i \ge 0 $, choose $ p_i $ to maximize $ f_i(p) $; if $ d_i < 0 $, choose $ p_i $ to minimize $ f_i(p) $.)
Thus, define for each $ i $:
$$
\Delta_i =
\begin{cases}
\max_{p \in \{1,\dots,n\}} f_i(p) & \text{if } d_i \ge 0 \\
\min_{p \in \{1,\dots,n\}} f_i(p) & \text{if } d_i < 0
\end{cases}
$$
Then the maximal arithmetic mean is:
$$
\frac{1}{n} \left( \sum_{i=1}^m x_i + \sum_{i=1}^m d_i \cdot \Delta_i \right)
$$
**Final Objective**
Compute:
$$
\boxed{ \frac{1}{n} \left( \sum_{i=1}^m x_i + \sum_{i=1}^m d_i \cdot \Delta_i \right) }
$$
where $ \Delta_i = \begin{cases} \max_p \sum_{j=1}^n |p - j| & d_i \ge 0 \\ \min_p \sum_{j=1}^n |p - j| & d_i < 0 \end{cases} $, and $ p \in \{1, \dots, n\} $.