#include <bits/stdc++.h>
using namespace std;
void addmod(int &a, long long b) {
  a = (a + b);
  if (a >= 1000000007) a -= 1000000007;
}
void mulmod(int &a, long long b) { a = (a * b) % 1000000007; }
int gi() {
  int a;
  scanf("%d", &a);
  return a;
}
long long gll() {
  long long a;
  scanf("%I64d", &a);
  return a;
}
char s[100007];
void doit() {
  scanf("%s", s);
  int n = strlen(s);
  int ix = 0;
  while (ix < n && s[ix] == 'a') ix++;
  if (ix == n) {
    s[n - 1] = 'z';
    return;
  }
  while (ix < n && s[ix] != 'a') {
    s[ix]--;
    ix++;
  }
}
int main() {
  doit();
  printf("%s\n", s);
  return 0;
}

### Prompt

Generate 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;
void addmod(int &a, long long b) {
  a = (a + b);
  if (a >= 1000000007) a -= 1000000007;
}
void mulmod(int &a, long long b) { a = (a * b) % 1000000007; }
int gi() {
  int a;
  scanf("%d", &a);
  return a;
}
long long gll() {
  long long a;
  scanf("%I64d", &a);
  return a;
}
char s[100007];
void doit() {
  scanf("%s", s);
  int n = strlen(s);
  int ix = 0;
  while (ix < n && s[ix] == 'a') ix++;
  if (ix == n) {
    s[n - 1] = 'z';
    return;
  }
  while (ix < n && s[ix] != 'a') {
    s[ix]--;
    ix++;
  }
}
int main() {
  doit();
  printf("%s\n", s);
  return 0;
}

```