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.
1
abbabababbabababbababbbabbbbabbbabaabaabba
 4
b
aabbbabaa
abbb
abbaab
 1
aaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbab

abbabababbabababbababbbabbbbabbbabaabaabba
 b
aabbbabaa
bbbb
bbbaab
 baaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbab

#include <bits/stdc++.h>
using namespace std;
int MOD = 1e9 + 7;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a = a % MOD;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % MOD;
      b--;
    }
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
long long int fermat_inv(long long int y) { return power(y, MOD - 2); }
long long int gcd(long long int a, long long int b) {
  return (b == 0) ? a : gcd(b, a % b);
}
long long int min(long long int a, long long int b) { return (a > b) ? b : a; }
long long int max(long long int a, long long int b) { return (a > b) ? a : b; }
bool prime[1000001];
vector<int> primes;
void SieveOfEratosthenes(int n) {
  memset(prime, true, sizeof(prime));
  prime[0] = prime[1] = 0;
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p < 1000001; p++)
    if (prime[p]) primes.push_back(p);
}
long long int fact[1000010];
long long int finv[1000010];
void factorial(int n) {
  fact[0] = 1;
  finv[0] = 1;
  for (int i = 1; i <= n; i++)
    fact[i] = fact[i - 1] * i, fact[i] %= MOD, finv[i] = fermat_inv(fact[i]);
}
long long int ncr(long long int n, long long int r) {
  if (n < r)
    return 0;
  else {
    long long int x = finv[r] * finv[n - r] % MOD;
    return fact[n] * x % MOD;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int te = 1;
  cin >> te;
  while (te--) {
    string s;
    cin >> s;
    int n = s.size();
    s[n - 1] = s[0];
    cout << s << '\n';
  }
}
