#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  int f = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] != 'a') f = 1;
  }
  if (f == 0) {
    for (int i = 0; i < s.size() - 1; i++) cout << "a";
    cout << "z";
    return 0;
  }
  f = 0;
  for (int i = 0; i < s.size(); i++) {
    if (f == 0 && s[i] == 'a') {
      cout << "a";
      continue;
    }
    if (s[i] != 'a') {
      f = 1;
      cout << (char)(s[i] - 1);
    }
    if (f == 1 && s[i] == 'a') {
      for (int j = i; j < s.size(); j++) cout << s[j];
      return 0;
    }
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  int f = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] != 'a') f = 1;
  }
  if (f == 0) {
    for (int i = 0; i < s.size() - 1; i++) cout << "a";
    cout << "z";
    return 0;
  }
  f = 0;
  for (int i = 0; i < s.size(); i++) {
    if (f == 0 && s[i] == 'a') {
      cout << "a";
      continue;
    }
    if (s[i] != 'a') {
      f = 1;
      cout << (char)(s[i] - 1);
    }
    if (f == 1 && s[i] == 'a') {
      for (int j = i; j < s.size(); j++) cout << s[j];
      return 0;
    }
  }
  return 0;
}

```