#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-15, PI = acos(-1);
const int MOD = int(1e9) + 7, MX = int(1e3) + 1;
void solve() {
  string s;
  cin >> s;
  int f = 0, done = 0;
  for (int i = 0; i < int(s.size()); i++) {
    int idx = s[i] - 'a' - 1 + 26;
    idx %= 26;
    char c = char(idx + 'a');
    if (c < s[i]) {
      s[i] = c, f = 1, done = 1;
    } else {
      if (f) break;
    }
  }
  if (!done)
    s[int(s.size()) - 1] =
        char(((s[int(s.size()) - 1] - 'a' - 1 + 26) % 26) + 'a');
  cout << s << "\n";
}
int main() { solve(); }

### Prompt

Create a solution in Cpp for the following problem:

You are given a non-empty string s consisting of lowercase English letters. You have to pick exactly one non-empty substring of s and shift all its letters 'z' <image> 'y' <image> 'x' <image> 'b' <image> 'a' <image> 'z'. In other words, each character is replaced with the previous character of English alphabet and 'a' is replaced with 'z'.

What is the lexicographically minimum string that can be obtained from s by performing this shift exactly once?

Input

The only line of the input contains the string s (1 ≤ |s| ≤ 100 000) consisting of lowercase English letters.

Output

Print the lexicographically minimum string that can be obtained from s by shifting letters of exactly one non-empty substring.

Examples

Input

codeforces


Output

bncdenqbdr


Input

abacaba


Output

aaacaba

Note

String s is lexicographically smaller than some other string t of the same length if there exists some 1 ≤ i ≤ |s|, such that s1 = t1, s2 = t2, ..., si - 1 = ti - 1, and si < ti.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-15, PI = acos(-1);
const int MOD = int(1e9) + 7, MX = int(1e3) + 1;
void solve() {
  string s;
  cin >> s;
  int f = 0, done = 0;
  for (int i = 0; i < int(s.size()); i++) {
    int idx = s[i] - 'a' - 1 + 26;
    idx %= 26;
    char c = char(idx + 'a');
    if (c < s[i]) {
      s[i] = c, f = 1, done = 1;
    } else {
      if (f) break;
    }
  }
  if (!done)
    s[int(s.size()) - 1] =
        char(((s[int(s.size()) - 1] - 'a' - 1 + 26) % 26) + 'a');
  cout << s << "\n";
}
int main() { solve(); }

```