#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <unordered_set>
#include <unordered_map>
#include <numeric>
#include <cmath>
// Dla macOs zachowaj includy w przeciwnym wypadku zastąp "#include <bits/stdc++.h> "
// szybki kod
#define all(v) (v).begin(), (v).end()
#define rep(i, a, b) for(int i = a;i <= b; i++)
#define per(i, a, b) for(int i = a;i >= b; i--)
#define pb push_back
#define ins insert
#define st first
#define nd second;
#define test int tc; cin>>tc; while(tc--)
// struktury danych
#define smldi set<map<long double, int > >
#define spumldidsi set<pair<unordered_map<long double, int>, deque<set<long long> > > >
// funkcje wspomagajace debugowanie programu
#define printv(a) { for(auto u : a) cout<<u<<" "; cout<<"/n"; }
#define debug(x) cerr << #x << " = " << x << endl;
// usingi
using namespace std;
using ll = long long;
using pii = pair<int,int>;
using vi = vector<int>;
using si = set<int>;
using mii = map<int,int>;
using bigi = __int128;
const ll p = 29, MOD = 1000000007;
ll hashe[1000005];
ll pot[1000005];
string s;
int n;
ll hash_(int a,int b){
if(a == 0) return hashe[b];
ll h1 = hashe[b], h2 = hashe[a-1];
//cout<<" mamy 2 hashe "<<h1<<" "<<h2<<" i 2 mnozemy razy "<<p<<" do "<<b-a + 1<<endl;
return (h1 - (h2 * pot[b - a + 1]) + (MOD * MOD)) % MOD;
}
void hashuj(){
hashe[0] = s[0];
//cout<<" pierwszy to "<<hashe[0]<<endl;
for(int i = 1;i < n;i++){
hashe[i] = (hashe[i - 1] * p + s[i]) % MOD;
//cout<<" ustawiamy nowy hash na poprzedni razy "<<p<<" "<<hashe[i]<<endl;
}
}
bool czy_te_same(int a,int b,int c,int d){
return hash_(a,b) == hash_(c,d);
}
void solve(){
cin>>s;
int akt = 0, wyn = 0;
n = s.size();
hashuj();
//cout<<hash_(0,1)<<endl<<hash_(2,3)<<endl;
for(int i=0;i < n/2;i++){
if(czy_te_same(akt, i, n - i - 1, n - akt - 1)){
//cout<<" znalezlismy te same na "<<akt<<" "<<i<<endl;
akt = i+1;
wyn+= 2;
}
}
if(akt != n/2 + n%2){
wyn++;
}
cout<<wyn<<endl;
}
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
pot[0] = 1;
for(int i=1;i<1000000;i++){
pot[i] = (pot[i-1] * p) % MOD;
}
test{
solve();
}
return 0;
}
