#include <bits/stdc++.h>
using namespace std;
const long long INF = 2000000000;
const long long mod = 1000000007;
char a[100100];
int main() {
  scanf("%s", a);
  int n = strlen(a);
  int s = 0;
  while (a[s] == 'a') {
    if (s == n - 1) break;
    s++;
  }
  int e = n;
  for (int i = s + 1; i < n; i++) {
    if (a[i] == 'a') {
      e = i;
      break;
    }
  }
  for (int i = 0; i < n; i++) {
    if (s <= i && i < e)
      printf("%c", (a[i] - 'a' + 25) % 26 + 'a');
    else
      printf("%c", a[i]);
  }
}

### Prompt

Your challenge is to write a Cpp solution 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 long long INF = 2000000000;
const long long mod = 1000000007;
char a[100100];
int main() {
  scanf("%s", a);
  int n = strlen(a);
  int s = 0;
  while (a[s] == 'a') {
    if (s == n - 1) break;
    s++;
  }
  int e = n;
  for (int i = s + 1; i < n; i++) {
    if (a[i] == 'a') {
      e = i;
      break;
    }
  }
  for (int i = 0; i < n; i++) {
    if (s <= i && i < e)
      printf("%c", (a[i] - 'a' + 25) % 26 + 'a');
    else
      printf("%c", a[i]);
  }
}

```