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?
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.
2 3
-1 3
0 0
-1 -4
-2.500000000000000
3 2
0 2
5 0
7.000000000000000
8 8
-21 -60
-96 -10
-4 -19
-27 -4
57 -15
-95 62
-42 1
-17 64
-16.500000000000000
1 1
0 0
0.000000000000000
100000 1
1000 1000
50000500.000000000000000
11 1
0 -10
-27.272727272727273
3 1
1 -1
0.333333333333333
1 2
-1 -1
-2 -2
-3.000000000000000
1 2
0 -1
0 1
0.000000000000000
1 1
1 -2
1.000000000000000
3 1
2 -1
1.333333333333333
3 1
0 -1
-0.666666666666667
1 1
-1000 -1000
-1000.000000000000000
1 1
0 -5
0.000000000000000
15 3
2 0
2 -5
-2 5
18.333333333333332
9 1
0 -5
-11.111111111111111
7 1
0 -1
-1.714285714285714
3 1
-2 -2
-3.333333333333333
3 1
5 -5
1.666666666666667
1 1
-1 -1
-1.000000000000000
7 1
-1 -5
-9.571428571428571
3 2
-2 -2
-2 -2
-6.666666666666667
5 1
0 -4
-4.800000000000000
5 1
-1 -5
-7.000000000000000
5 1
0 -2
-2.400000000000000
3 5
1 -1000
1 -1000
1 -1000
1 -1000
1 -1000
-3328.333333333333485
1 1
0 -1
0.000000000000000
1 2
0 -3
0 -3
0.000000000000000
7 1
2 -3
-3.142857142857143
3 2
-1 -1
-1 -1
-3.333333333333333
5 1
-1 -162
-195.400000000000006
5 10
-506 -243
727 -141
-548 -306
740 880
-744 -116
-84 182
-859 -108
64 86
135 446
69 -184
864.399999999999977
5 1
0 -1
-1.200000000000000
5 12
634 895
143 730
901 245
386 486
395 -111
-469 -104
-681 -623
-900 843
889 -883
476 -304
777 986
206 -491
8107.800000000000182
3 3
4 2
5 0
6 -1
16.333333333333332
1 3
4 2
5 0
6 -1
15.000000000000000
85 10
-223 435
-771 455
72 -940
490 -178
400 -117
169 -527
836 610
849 944
572 -237
-428 -428
53047.388235294114565
69 10
-8 4
-3 3
7 5
5 -9
8 1
7 -5
-8 -8
9 3
1 1
0 6
420.579710144927560
1 10
1 1
1 0
1 0
1 0
-1 0
0 1
1 0
0 0
2 1
9 2
15.000000000000000
5 4
0 1
0 2
0 3
0 -9
1.200000000000000
import sys
input = sys.stdin.readline


def f(u):
    return u*(u+1)//2


n, m = map(int, input().split())
x1, x2 = n//2, (n-1)//2
x, y, c, d = f(x1) + f(x2), f(n-1), 0, 0
for i in range(m):
    a, b = map(int, input().split())
    c += a
    d += max(b*x, b*y)
print((c*n+d)/n)