/*       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;}