#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int advceil(long long int num, long long int den) {
  return (num % den == 0 ? num / den : num / den + 1);
}
long long int lstbt(long long int val) {
  long long int msk = val & (val - 1);
  return log2(val ^ msk);
}
long long int modmul(long long int a, long long int b) {
  if (a < 1e9 and b < 1e9) return (a * b) % 1000000007;
  a %= 1000000007;
  b %= 1000000007;
  long long int res = 0;
  while (b > 0) {
    if (b & 1) res = (res + a) % 1000000007;
    a = (a * 2) % 1000000007;
    b /= 2;
  }
  return res;
}
long long int modexpo(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = modmul(res, a);
    a = modmul(a, a);
    b /= 2;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return a == 0 ? b : gcd(b % a, a);
}
vector<long long int> CALCfactor(long long int n) {
  vector<long long int> factor(n + 2, 0);
  for (long long int i = 4; i <= n; i += 2) factor[i] = 2;
  for (long long int j = 3; j <= n; j += 2) {
    if (factor[j]) continue;
    for (long long int i = 2 * j; i <= n; i += j) factor[i] = j;
  }
  return factor;
}
vector<long long int> CALCprimeNUM(long long int n) {
  vector<long long int> factor = CALCfactor(n);
  vector<long long int> primeNUM;
  primeNUM.reserve(n + 2);
  for (long long int i = 2; i <= n; i++)
    if (!factor[i]) primeNUM.push_back(i);
  return primeNUM;
}
vector<long long int> CALCprimeFACTOR(long long int n) {
  vector<long long int> factor = CALCfactor(n);
  vector<long long int> ans;
  while (factor[n] != 0) {
    ans.push_back(factor[n]);
    n /= factor[n];
  }
  if (n > 1) ans.push_back(n);
  return ans;
}
vector<long long int> unique(vector<long long int> x) {
  sort(x.begin(), x.end());
  set<long long int> s;
  vector<long long int> ans;
  ans.reserve(x.size());
  for (auto elem : x) s.insert(elem);
  for (auto elem : s) ans.push_back(elem);
  return ans;
}
pair<vector<long long int>, vector<long long int> > getFact(long long int n) {
  vector<long long int> fact(n + 1, 1), invfact(n + 1, 1);
  for (long long int i = 1; i <= n; i++)
    fact[i] = (i * (fact[i - 1])) % 1000000007;
  for (long long int i = 1; i <= n; i++)
    invfact[i] = (modexpo(i, 1000000007 - 2) * invfact[i - 1]) % 1000000007;
  return {fact, invfact};
}
void compress(vector<long long int>& arr, long long int n) {
  map<long long int, vector<long long int> > pos;
  for (long long int i = 1; i <= n; i++) pos[arr[i]].push_back(i);
  long long int cnt = 1;
  for (auto itr : pos) {
    for (auto elem : itr.second) arr[elem] = cnt;
    cnt++;
  }
}
bool rcomp(long long int a, long long int b) { return a > b; }
vector<long long int> prefix(string pat, string s) {
  long long int len = s.size();
  s = pat + '$' + s;
  vector<long long int> dp(s.size(), 0);
  long long int L = 0;
  long long int R = 0;
  for (long long int i = 1; i < s.size(); i++) {
    if (i > R) {
      L = i, R = i;
      while (R < s.size() and s[R] == s[R - L]) R++;
      R--;
      dp[i] = R - L + 1;
    } else {
      if (dp[i - L] < R - i + 1)
        dp[i] = dp[i - L];
      else {
        L = i;
        while (R < s.size() and s[R] == s[R - L]) R++;
        R--;
        dp[i] = R - L + 1;
      }
    }
  }
  vector<long long int> ans(len, 0);
  for (long long int i = pat.length() + 1; i < s.size(); i++)
    ans[i - pat.length() - 1] = dp[i];
  return ans;
}
void solve() {
  string s;
  cin >> s;
  long long int lo = 0;
  long long int hi = s.size() - 1;
  while (lo <= hi and s[lo] == s[hi]) lo++, hi--;
  if (lo > hi) {
    cout << s << "\n";
    return;
  }
  string longestpalindrom = string(1, s[lo]);
  string pat = s.substr(lo, hi - lo + 1);
  string rev = pat;
  reverse(rev.begin(), rev.end());
  vector<long long int> dp1 = prefix(pat, rev);
  for (long long int i = 0; i < dp1.size(); i++) {
    if (dp1[i] == hi - i - lo + 1) {
      if (longestpalindrom.size() < hi - i - lo + 1) {
        longestpalindrom = s.substr(lo, hi - i - lo + 1);
      }
      break;
    }
  }
  vector<long long int> dp2 = prefix(rev, pat);
  for (long long int i = 0; i < dp2.size(); i++) {
    if (dp2[i] == hi - (lo + i) + 1) {
      if (longestpalindrom.size() < hi - i - lo + 1) {
        longestpalindrom = s.substr(lo + i, hi - i - lo + 1);
      }
      break;
    }
  }
  cout << s.substr(0, lo) + longestpalindrom + s.substr(hi + 1, lo) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  auto start1 = high_resolution_clock::now();
  long long int t;
  cin >> t;
  while (t--) solve();
  auto stop1 = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop1 - start1);
}

### Prompt

Please create a solution in cpp to the following problem:

This is the easy version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.

You are given a string s, consisting of lowercase English letters. Find the longest string, t, which satisfies the following conditions: 

  * The length of t does not exceed the length of s. 
  * t is a palindrome. 
  * There exists two strings a and b (possibly empty), such that t = a + b ( "+" represents concatenation), and a is prefix of s while b is suffix of s. 

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000), the number of test cases. The next t lines each describe a test case.

Each test case is a non-empty string s, consisting of lowercase English letters.

It is guaranteed that the sum of lengths of strings over all test cases does not exceed 5000.

Output

For each test case, print the longest string which satisfies the conditions described above. If there exists multiple possible solutions, print any of them.

Example

Input


5
a
abcdfdcecba
abbaxyzyx
codeforces
acbba


Output


a
abcdfdcba
xyzyx
c
abba

Note

In the first test, the string s = "a" satisfies all conditions.

In the second test, the string "abcdfdcba" satisfies all conditions, because:

  * Its length is 9, which does not exceed the length of the string s, which equals 11. 
  * It is a palindrome. 
  * "abcdfdcba" = "abcdfdc" + "ba", and "abcdfdc" is a prefix of s while "ba" is a suffix of s. 



It can be proven that there does not exist a longer string which satisfies the conditions.

In the fourth test, the string "c" is correct, because "c" = "c" + "" and a or b can be empty. The other possible solution for this test is "s".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int advceil(long long int num, long long int den) {
  return (num % den == 0 ? num / den : num / den + 1);
}
long long int lstbt(long long int val) {
  long long int msk = val & (val - 1);
  return log2(val ^ msk);
}
long long int modmul(long long int a, long long int b) {
  if (a < 1e9 and b < 1e9) return (a * b) % 1000000007;
  a %= 1000000007;
  b %= 1000000007;
  long long int res = 0;
  while (b > 0) {
    if (b & 1) res = (res + a) % 1000000007;
    a = (a * 2) % 1000000007;
    b /= 2;
  }
  return res;
}
long long int modexpo(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = modmul(res, a);
    a = modmul(a, a);
    b /= 2;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return a == 0 ? b : gcd(b % a, a);
}
vector<long long int> CALCfactor(long long int n) {
  vector<long long int> factor(n + 2, 0);
  for (long long int i = 4; i <= n; i += 2) factor[i] = 2;
  for (long long int j = 3; j <= n; j += 2) {
    if (factor[j]) continue;
    for (long long int i = 2 * j; i <= n; i += j) factor[i] = j;
  }
  return factor;
}
vector<long long int> CALCprimeNUM(long long int n) {
  vector<long long int> factor = CALCfactor(n);
  vector<long long int> primeNUM;
  primeNUM.reserve(n + 2);
  for (long long int i = 2; i <= n; i++)
    if (!factor[i]) primeNUM.push_back(i);
  return primeNUM;
}
vector<long long int> CALCprimeFACTOR(long long int n) {
  vector<long long int> factor = CALCfactor(n);
  vector<long long int> ans;
  while (factor[n] != 0) {
    ans.push_back(factor[n]);
    n /= factor[n];
  }
  if (n > 1) ans.push_back(n);
  return ans;
}
vector<long long int> unique(vector<long long int> x) {
  sort(x.begin(), x.end());
  set<long long int> s;
  vector<long long int> ans;
  ans.reserve(x.size());
  for (auto elem : x) s.insert(elem);
  for (auto elem : s) ans.push_back(elem);
  return ans;
}
pair<vector<long long int>, vector<long long int> > getFact(long long int n) {
  vector<long long int> fact(n + 1, 1), invfact(n + 1, 1);
  for (long long int i = 1; i <= n; i++)
    fact[i] = (i * (fact[i - 1])) % 1000000007;
  for (long long int i = 1; i <= n; i++)
    invfact[i] = (modexpo(i, 1000000007 - 2) * invfact[i - 1]) % 1000000007;
  return {fact, invfact};
}
void compress(vector<long long int>& arr, long long int n) {
  map<long long int, vector<long long int> > pos;
  for (long long int i = 1; i <= n; i++) pos[arr[i]].push_back(i);
  long long int cnt = 1;
  for (auto itr : pos) {
    for (auto elem : itr.second) arr[elem] = cnt;
    cnt++;
  }
}
bool rcomp(long long int a, long long int b) { return a > b; }
vector<long long int> prefix(string pat, string s) {
  long long int len = s.size();
  s = pat + '$' + s;
  vector<long long int> dp(s.size(), 0);
  long long int L = 0;
  long long int R = 0;
  for (long long int i = 1; i < s.size(); i++) {
    if (i > R) {
      L = i, R = i;
      while (R < s.size() and s[R] == s[R - L]) R++;
      R--;
      dp[i] = R - L + 1;
    } else {
      if (dp[i - L] < R - i + 1)
        dp[i] = dp[i - L];
      else {
        L = i;
        while (R < s.size() and s[R] == s[R - L]) R++;
        R--;
        dp[i] = R - L + 1;
      }
    }
  }
  vector<long long int> ans(len, 0);
  for (long long int i = pat.length() + 1; i < s.size(); i++)
    ans[i - pat.length() - 1] = dp[i];
  return ans;
}
void solve() {
  string s;
  cin >> s;
  long long int lo = 0;
  long long int hi = s.size() - 1;
  while (lo <= hi and s[lo] == s[hi]) lo++, hi--;
  if (lo > hi) {
    cout << s << "\n";
    return;
  }
  string longestpalindrom = string(1, s[lo]);
  string pat = s.substr(lo, hi - lo + 1);
  string rev = pat;
  reverse(rev.begin(), rev.end());
  vector<long long int> dp1 = prefix(pat, rev);
  for (long long int i = 0; i < dp1.size(); i++) {
    if (dp1[i] == hi - i - lo + 1) {
      if (longestpalindrom.size() < hi - i - lo + 1) {
        longestpalindrom = s.substr(lo, hi - i - lo + 1);
      }
      break;
    }
  }
  vector<long long int> dp2 = prefix(rev, pat);
  for (long long int i = 0; i < dp2.size(); i++) {
    if (dp2[i] == hi - (lo + i) + 1) {
      if (longestpalindrom.size() < hi - i - lo + 1) {
        longestpalindrom = s.substr(lo + i, hi - i - lo + 1);
      }
      break;
    }
  }
  cout << s.substr(0, lo) + longestpalindrom + s.substr(hi + 1, lo) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  auto start1 = high_resolution_clock::now();
  long long int t;
  cin >> t;
  while (t--) solve();
  auto stop1 = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop1 - start1);
}

```