#include <bits/stdc++.h>
using namespace std;
void setIO(string name = "") {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (int((name).size())) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long mcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool prime(long long n, long long i) {
  if (i > sqrt(n)) return true;
  if (n % i == 0)
    return false;
  else
    return prime(n, i + 1);
}
struct compii {
  bool operator()(const pair<int, int>& a, const pair<int, int>& b) {
    if (a.first == a.second) return a.second < b.second;
    return a.first > b.first;
  }
};
bool comp(int a, int b) { return a > b; }
namespace operators {
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
  in >> x.first >> x.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> x) {
  out << x.first << " " << x.second;
  return out;
}
template <typename T1>
istream& operator>>(istream& in, vector<T1>& x) {
  for (auto& i : x) in >> i;
  return in;
}
template <typename T1>
ostream& operator<<(ostream& out, vector<T1>& x) {
  for (auto& i : x) out << i << " ";
  return out;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, vector<pair<T1, T2>>& x) {
  for (auto& i : x) out << i.first << " " << i.second;
  return out;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, vector<pair<T1, T2>>& x) {
  for (auto& i : x) in >> i.first >> i.second;
  return in;
}
}  // namespace operators
using namespace operators;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
const int mod = 1e9 + 7;
const long long inf = 1e18;
void solve() {
  string second;
  cin >> second;
  int n = int((second).size());
  int c1 = 0, c2 = 0;
  for (int i = 0; i < n - 1; i++) {
    string aux = "";
    aux += second[i];
    aux += second[i + 1];
    if (aux == "ab") c1++;
  }
  for (int i = 0; i < n - 1; i++) {
    string aux = "";
    aux += second[i];
    aux += second[i + 1];
    if (aux == "ba") c2++;
  }
  if (c1 == c2) {
    cout << second << endl;
    return;
  } else {
    second[0] = 'b';
    second[n - 1] = 'b';
    cout << second << endl;
  }
}
int main() {
  setIO("");
  int t = 1;
  cin >> t;
  while (t-- > 0) solve();
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

You are given a string s of length n consisting of characters a and/or b.

Let \operatorname{AB}(s) be the number of occurrences of string ab in s as a substring. Analogically, \operatorname{BA}(s) is the number of occurrences of ba in s as a substring.

In one step, you can choose any index i and replace s_i with character a or b.

What is the minimum number of steps you need to make to achieve \operatorname{AB}(s) = \operatorname{BA}(s)?

Reminder:

The number of occurrences of string d in s as substring is the number of indices i (1 ≤ i ≤ |s| - |d| + 1) such that substring s_i s_{i + 1} ... s_{i + |d| - 1} is equal to d. For example, \operatorname{AB}(aabbbabaa) = 2 since there are two indices i: i = 2 where aabbbabaa and i = 6 where aabbbabaa.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.

The first and only line of each test case contains a single string s (1 ≤ |s| ≤ 100, where |s| is the length of the string s), consisting only of characters a and/or b.

Output

For each test case, print the resulting string s with \operatorname{AB}(s) = \operatorname{BA}(s) you'll get making the minimum number of steps.

If there are multiple answers, print any of them.

Example

Input


4
b
aabbbabaa
abbb
abbaab


Output


b
aabbbabaa
bbbb
abbaaa

Note

In the first test case, both \operatorname{AB}(s) = 0 and \operatorname{BA}(s) = 0 (there are no occurrences of ab (ba) in b), so can leave s untouched.

In the second test case, \operatorname{AB}(s) = 2 and \operatorname{BA}(s) = 2, so you can leave s untouched. 

In the third test case, \operatorname{AB}(s) = 1 and \operatorname{BA}(s) = 0. For example, we can change s_1 to b and make both values zero.

In the fourth test case, \operatorname{AB}(s) = 2 and \operatorname{BA}(s) = 1. For example, we can change s_6 to a and make both values equal to 1.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void setIO(string name = "") {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (int((name).size())) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long mcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool prime(long long n, long long i) {
  if (i > sqrt(n)) return true;
  if (n % i == 0)
    return false;
  else
    return prime(n, i + 1);
}
struct compii {
  bool operator()(const pair<int, int>& a, const pair<int, int>& b) {
    if (a.first == a.second) return a.second < b.second;
    return a.first > b.first;
  }
};
bool comp(int a, int b) { return a > b; }
namespace operators {
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
  in >> x.first >> x.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> x) {
  out << x.first << " " << x.second;
  return out;
}
template <typename T1>
istream& operator>>(istream& in, vector<T1>& x) {
  for (auto& i : x) in >> i;
  return in;
}
template <typename T1>
ostream& operator<<(ostream& out, vector<T1>& x) {
  for (auto& i : x) out << i << " ";
  return out;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, vector<pair<T1, T2>>& x) {
  for (auto& i : x) out << i.first << " " << i.second;
  return out;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, vector<pair<T1, T2>>& x) {
  for (auto& i : x) in >> i.first >> i.second;
  return in;
}
}  // namespace operators
using namespace operators;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
const int mod = 1e9 + 7;
const long long inf = 1e18;
void solve() {
  string second;
  cin >> second;
  int n = int((second).size());
  int c1 = 0, c2 = 0;
  for (int i = 0; i < n - 1; i++) {
    string aux = "";
    aux += second[i];
    aux += second[i + 1];
    if (aux == "ab") c1++;
  }
  for (int i = 0; i < n - 1; i++) {
    string aux = "";
    aux += second[i];
    aux += second[i + 1];
    if (aux == "ba") c2++;
  }
  if (c1 == c2) {
    cout << second << endl;
    return;
  } else {
    second[0] = 'b';
    second[n - 1] = 'b';
    cout << second << endl;
  }
}
int main() {
  setIO("");
  int t = 1;
  cin >> t;
  while (t-- > 0) solve();
  return 0;
}

```