Problem: This is the hard 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 ≤ 10^5), 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 10^6.

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".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B, typename C>
struct triple {
  A X;
  B Y;
  C Z;
  triple(A a = 0, B b = 0, C c = 0) : X(a), Y(b), Z(c) {}
};
template <typename A, typename B, typename C>
triple<A, B, C> make_triple(A a = 0, B b = 0, C c = 0) {
  return triple<A, B, C>(a, b, c);
}
template <typename A, typename B, typename C>
bool operator<(const triple<A, B, C>& a, const triple<A, B, C>& b) {
  if (a.X != b.X) return a.X < b.X;
  if (a.Y != b.Y) return a.Y < b.Y;
  return a.Z < b.Z;
}
template <typename T, typename SS>
ostream& operator<<(ostream& ofs, const pair<T, SS>& p) {
  ofs << "( " << p.first << " , " << p.second << " )";
  return ofs;
}
template <typename T>
void print(T a) {
  for (auto i : a) cout << i << ' ';
  cout << '\n';
}
template <typename T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
template <typename T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long gen_base(const long long before, const long long after) {
  auto seed =
      std::chrono::high_resolution_clock::now().time_since_epoch().count();
  std::mt19937 mt_rand(seed);
  long long base =
      std::uniform_int_distribution<long long>(before + 1, after)(mt_rand);
  return base % 2 == 0 ? base - 1 : base;
}
struct PolyHash {
  static const long long mod = (long long)1e9 + 123;
  static std::vector<long long> pow1;
  static std::vector<unsigned long long> pow2;
  static long long base;
  static inline long long diff(long long a, long long b) {
    return (a -= b) < 0 ? a + mod : a;
  }
  std::vector<long long> pref1;
  std::vector<unsigned long long> pref2;
  PolyHash(const std::string& s)
      : pref1(s.size() + 1u, 0), pref2(s.size() + 1u, 0) {
    const long long n = s.size();
    while ((long long)pow1.size() <= n) {
      pow1.push_back(1LL * pow1.back() * base % mod);
      pow2.push_back(pow2.back() * base);
    }
    for (long long i = 0; i < n; ++i) {
      pref1[i + 1] = (pref1[i] + 1LL * s[i] * pow1[i]) % mod;
      pref2[i + 1] = pref2[i] + s[i] * pow2[i];
    }
  }
  inline std::pair<long long, unsigned long long> operator()(
      const long long pos, const long long len,
      const long long mxPow = 0) const {
    long long hash1 = pref1[pos + len] - pref1[pos];
    unsigned long long hash2 = pref2[pos + len] - pref2[pos];
    if (hash1 < 0) hash1 += mod;
    if (mxPow != 0) {
      hash1 = 1LL * hash1 * pow1[mxPow - (pos + len - 1)] % mod;
      hash2 *= pow2[mxPow - (pos + len - 1)];
    }
    return std::make_pair(hash1, hash2);
  }
};
long long PolyHash::base((long long)1e9 + 7);
std::vector<long long> PolyHash::pow1{1};
std::vector<unsigned long long> PolyHash::pow2{1};
signed main() {
  srand(time(NULL));
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  PolyHash::base = gen_base(256, PolyHash::mod);
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long n = s.size();
    string a = s;
    string b = "";
    reverse_copy(s.begin(), s.end(), back_inserter(b));
    string ans;
    long long i = 0;
    while (i < n && a[i] == b[i]) ans.push_back(a[i++]);
    if (i == n) {
      cout << ans << '\n';
      continue;
    }
    auto fnd = [](string s) {
      long long n = (long long)(s.size());
      string rs;
      reverse_copy(s.begin(), s.end(), back_inserter(rs));
      const long long mxPow = n;
      PolyHash a(s), b(rs);
      auto good = [&](long long x, long long ln) {
        long long y = x + ln - 1;
        return a(x, ln, mxPow) == b(n - y - 1, ln, mxPow);
      };
      long long p1, p2;
      p1 = p2 = -1;
      for (long long i = 0; i + i < n; i++) {
        if (good(0, 2 * i + 1)) p1 = 2 * i + 1;
      }
      for (long long i = 0; 2 * i + 1 < n; i++) {
        if (good(0, 2 * i + 2)) p2 = 2 * i + 2;
      }
      return (p1 > p2) ? s.substr(0, p1) : s.substr(0, p2);
    };
    string c;
    for (long long j = i; j < n - i; j++) c.push_back(a[j]);
    string x = fnd(c);
    reverse(c.begin(), c.end());
    string y = fnd(c);
    string z = ((long long)(x.size()) > (long long)(y.size()) ? x : y);
    cout << ans << z;
    reverse(ans.begin(), ans.end());
    cout << ans << '\n';
  }
}