/*
实际上就是求一个k，满足k<=n/2，且gcd(n,k)=1
如果n为奇数，k为[n/2]
如果n为偶数，k=n/2-1-(n/2)%2
*/
#include <iostream>
using namespace std;
string s;
void div2() {
string t;
int l = s.size() - 1, tem = s[0] - '0';
if (tem > 1) t += '0' + tem / 2;
tem &= 1;
for (int i = 1; i <= l; i++) {
tem = tem * 10 + s[i] - '0';
t += '0' + tem / 2;
tem &= 1;
}
s = t;
}
void cut1() {
int t = s.size() - 1;
while (s[t] == '0')
s[t--] = '9';
s[t] = s[t] - 1;
}
int main() {
cin >> s;
int l = s.size() - 1;
if ( (s[l] - '0') & 1) {
cut1(); div2();
cout << s;
}
else {
div2();
l = s.size() - 1;
if ( (s[l] - '0') & 1) cut1();
cut1();
cout << s;
}
return 0;
}