#include <bits/stdc++.h>
//#pragma GCC optimize("Ofast","-funroll-loops")
//#pragma GCC target("sse4.1","sse4.2","ssse3","sse3","sse2","sse","avx2","avx","popcnt","tune=native")

using namespace std;

typedef long double ldb;
typedef long long i64;
typedef unsigned long long u64;
typedef unsigned int u32;
typedef pair<int, int> pii;

// priority_queue<int, vector<int>, greater<int>> minq;
// ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
// fflush(stdout);

const int inf = 1000000007;
const i64 prm = 998244353;
const i64 inf2 = ((i64)inf) * inf;
const i64 bone = 1;
const int maxn = 500010;

inline int read(){
  int x=0,f=0; char ch=getchar();
  while(!isdigit(ch)) f|=(ch==45),ch=getchar();
  while(isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
  return f?-x:x;
}

/*
   1. 请再三确认题意
   2. 对于复杂的数据结构+贪心/结论/计数题，可以考虑先写暴力，确认想法正确，再逐步优化
*/

void once() {
  string s;
  cin >> s;
  int sz = s.size(), ans = 0;
  string res;
  for (int i = 0; i < sz; ++i) {
    for (int j = i+1; j < sz; ++j) {
      int k = min(j-i, sz-j);
      int len = 0;
      while (len < k && s[i+len] == s[j+len]) {
        len++;
      }
      if (len > ans) {
        res = s.substr(i, len);
        ans = len;
      }
    }
  }
  cout << res << endl;
}

int main() {
  int tes = 1;
  tes = read();
  while (tes--) {
    once();
  }
  return 0;
}