A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends.

There are *n* schools numerated from 1 to *n*. One can travel between each pair of them, to do so, he needs to buy a ticket. The ticker between schools *i* and *j* costs <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/48a4e5770c4d95918ab20e59965b1361f52acc70.png" style="max-width: 100.0%;max-height: 100.0%;"/> and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school.
The first line contains a single integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>) — the number of schools.
In the first example we can buy a ticket between the schools that costs <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c67d72010e0498bfd065a6a38fdeaec90358507b.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
2
0
10
4
43670
21834
4217
2108
17879
8939
31809
15904
40873
20436
77859
38929
53022
26510
79227
39613
100000
49999
82801
41400
5188
2593
86539
43269
12802
6400
20289
10144
32866
16432
33377
16688
31775
15887
60397
30198
100000
49999
99999
49999
99998
49998
99997
49998
99996
49997
1
0
2
0
3
1
4
1
1
0
3
1
n = int(input())
cost = 0
d = {1:1}
pos = 1
flag = True
pre = 1
k = 0
while k < n:
    if flag:
        pos = n + 1 - pos
        d[pos] = 1
        k += 1
        flag = False
    else:
        pos = pre + 1
        pre += 1
        k += 1
        if pos not in d:
            d[pos] = 1
            cost += 1
        flag = True

print(cost)