#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using pi = pair<ll, ll>;
void setIO(string name = "") {
  if ((int)(name).size()) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
}
ll gcd(ll a, ll b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a > b) return gcd(a % b, b);
  return gcd(a, b % a);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    string a;
    cin >> a;
    int n = a.size();
    string b = a, c = a, d = a;
    b[0] = (a[0] == 'a') ? 'b' : 'a';
    c[n - 1] = (a[n - 1] == 'b') ? 'a' : 'b';
    d[0] = (a[0] == 'a') ? 'b' : 'a';
    d[n - 1] = (a[n - 1] == 'b') ? 'a' : 'b';
    int sum = 0;
    for (int i = 0; i < n - 1; i++) {
      if (a[i] == 'a' && a[i + 1] == 'b') {
        sum++;
      } else if (a[i] == 'b' && a[i + 1] == 'a') {
        sum--;
      }
    }
    if (sum == 0) {
      cout << a << endl;
      continue;
    }
    sum = 0;
    for (int i = 0; i < n - 1; i++) {
      if (b[i] == 'a' && b[i + 1] == 'b') {
        sum++;
      } else if (b[i] == 'b' && b[i + 1] == 'a') {
        sum--;
      }
    }
    if (sum == 0) {
      cout << b << endl;
      continue;
    }
    sum = 0;
    for (int i = 0; i < n - 1; i++) {
      if (c[i] == 'a' && c[i + 1] == 'b') {
        sum++;
      } else if (c[i] == 'b' && c[i + 1] == 'a') {
        sum--;
      }
    }
    if (sum == 0) {
      cout << c << endl;
      continue;
    }
    sum = 0;
    for (int i = 0; i < n - 1; i++) {
      if (d[i] == 'a' && d[i + 1] == 'b') {
        sum++;
      } else if (d[i] == 'b' && d[i + 1] == 'a') {
        sum--;
      }
    }
    if (sum == 0) {
      cout << d << endl;
      continue;
    }
  }
}

### Prompt

Your task is to create 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;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using pi = pair<ll, ll>;
void setIO(string name = "") {
  if ((int)(name).size()) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
}
ll gcd(ll a, ll b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a > b) return gcd(a % b, b);
  return gcd(a, b % a);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    string a;
    cin >> a;
    int n = a.size();
    string b = a, c = a, d = a;
    b[0] = (a[0] == 'a') ? 'b' : 'a';
    c[n - 1] = (a[n - 1] == 'b') ? 'a' : 'b';
    d[0] = (a[0] == 'a') ? 'b' : 'a';
    d[n - 1] = (a[n - 1] == 'b') ? 'a' : 'b';
    int sum = 0;
    for (int i = 0; i < n - 1; i++) {
      if (a[i] == 'a' && a[i + 1] == 'b') {
        sum++;
      } else if (a[i] == 'b' && a[i + 1] == 'a') {
        sum--;
      }
    }
    if (sum == 0) {
      cout << a << endl;
      continue;
    }
    sum = 0;
    for (int i = 0; i < n - 1; i++) {
      if (b[i] == 'a' && b[i + 1] == 'b') {
        sum++;
      } else if (b[i] == 'b' && b[i + 1] == 'a') {
        sum--;
      }
    }
    if (sum == 0) {
      cout << b << endl;
      continue;
    }
    sum = 0;
    for (int i = 0; i < n - 1; i++) {
      if (c[i] == 'a' && c[i + 1] == 'b') {
        sum++;
      } else if (c[i] == 'b' && c[i + 1] == 'a') {
        sum--;
      }
    }
    if (sum == 0) {
      cout << c << endl;
      continue;
    }
    sum = 0;
    for (int i = 0; i < n - 1; i++) {
      if (d[i] == 'a' && d[i + 1] == 'b') {
        sum++;
      } else if (d[i] == 'b' && d[i + 1] == 'a') {
        sum--;
      }
    }
    if (sum == 0) {
      cout << d << endl;
      continue;
    }
  }
}

```