/*
复杂度O(n2)的简洁求法
int fast(char *p)
{
int ans = 1;
for (int i = 1; p[i]; ++i)
{
int s = i, e = i, t;
while (p[e + 1] == p[i]) ++e;
i = e;
while (p[s - 1] == p[e + 1]) --s, ++e;
if ((t = e - s + 1) > ans) ans = t;
}
return ans;
} */
#include <bits/stdc++.h>using namespace std;int ans[1000*2];string change(string a){ string b; b+='$'; b+='#'; for(int i=0;i<a.size();i++) { b+=a[i]; b+='#'; } return b;}void huiwen(string a){ int maxans=-1; int maxid=0; int id=0; memset(ans,0,sizeof(a)); string
b=change(a); for(int i=0;i<b.size();i++) { if(maxid>i) { maxid=ans[id*2-i]<maxid-i?ans[id*2-i]:maxid-i; } else { ans[i]=1; } while(b[i+ans[i]]==b[i-ans[i]])ans[i]++; if(ans[i]+i>maxid) { maxid=ans[i]+i; id=i; } if(maxans<ans[i]) maxans=ans[i]; } cout<<maxans-1<<endl;}int
main(){ string a; while(cin>>a) { huiwen(a); } return 0;}
