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>
long long mod = 1000000007;
using namespace std;
const int MAX_N = 1e6 + 2, BIT = 34;
long long poww(long long x, long long n) {
  if (n == 0) return 1;
  if (n == 1) return x % mod;
  long long ans = poww(x, n / 2);
  ans = (ans * ans) % mod;
  if (n % 2) return (ans * x) % mod;
  return ans % mod;
}
struct pt {
  long long x, y;
  pt() {}
  pt(long long _x, long long _y) : x(_x), y(_y) {}
  pt operator+(const pt &p) const { return pt(x + p.x, y + p.y); }
  pt operator-(const pt &p) const { return pt(x - p.x, y - p.y); }
  long long cross(const pt &p) const { return x * p.y - y * p.x; }
  long long dot(const pt &p) const { return x * p.x + y * p.y; }
  long long cross(const pt &a, const pt &b) const {
    return (a - *this).cross(b - *this);
  }
  long long dot(const pt &a, const pt &b) const {
    return (a - *this).dot(b - *this);
  }
  long long sqrLen() const { return this->dot(*this); }
  void Out() { cout << x << " " << y << endl; }
};
long long Gcd(long long a, long long b) {
  if (b > a) swap(a, b);
  if (b == 0) return a;
  return Gcd(b, a % b);
}
long long Pref[MAX_N];
long long Inv[MAX_N];
long long A[MAX_N];
int main() {
  long long Tc = 1;
  cin >> Tc;
  while (Tc--) {
    string s;
    cin >> s;
    long long n = s.size();
    char f = s[0];
    long long fi = 0;
    char e = s[n - 1];
    long long en = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == f)
        fi++;
      else
        break;
    }
    for (int i = n - 1; i >= 0; i--) {
      if (s[i] == e)
        en++;
      else
        break;
    }
    if (f != e) {
      s[0] = e;
    }
    cout << s << endl;
  }
  return 0;
}
