#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 123;
char s[N];
char prev(char c) { return c == 'a' ? 'z' : (char)(c - 1); }
int main() {
  scanf("%s", s);
  int n = strlen(s);
  int left = 0;
  while (left + 1 < n && s[left] == 'a') {
    left++;
  }
  int right = left;
  while (right + 1 < n && s[right + 1] != 'a') {
    right++;
  }
  for (int i = 0; i < n; i++) {
    if (left <= i && i <= right) {
      printf("%c", prev(s[i]));
    } else {
      printf("%c", s[i]);
    }
  }
  printf("\n");
  return 0;
}

### Prompt

Please create a solution in cpp to 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 int N = 1e5 + 123;
char s[N];
char prev(char c) { return c == 'a' ? 'z' : (char)(c - 1); }
int main() {
  scanf("%s", s);
  int n = strlen(s);
  int left = 0;
  while (left + 1 < n && s[left] == 'a') {
    left++;
  }
  int right = left;
  while (right + 1 < n && s[right + 1] != 'a') {
    right++;
  }
  for (int i = 0; i < n; i++) {
    if (left <= i && i <= right) {
      printf("%c", prev(s[i]));
    } else {
      printf("%c", s[i]);
    }
  }
  printf("\n");
  return 0;
}

```