#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1000000007;
void solve() {
  string c;
  cin >> c;
  if (*max_element((c).begin(), (c).end()) ==
          *min_element((c).begin(), (c).end()) &&
      *max_element((c).begin(), (c).end()) == 'a') {
    for (int i = 0; i < c.length() - 1; ++i) {
      cout << 'a';
    }
    cout << 'z';
    return;
  }
  int i = 0;
  while (i < c.length() && c[i] == 'a') i++;
  while (i < c.length() && c[i] != 'a') {
    c[i] = char(c[i] - 1);
    i++;
  }
  cout << c;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

### Prompt

Develop a solution in CPP to the problem described below:

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 int mod = 1000000007;
void solve() {
  string c;
  cin >> c;
  if (*max_element((c).begin(), (c).end()) ==
          *min_element((c).begin(), (c).end()) &&
      *max_element((c).begin(), (c).end()) == 'a') {
    for (int i = 0; i < c.length() - 1; ++i) {
      cout << 'a';
    }
    cout << 'z';
    return;
  }
  int i = 0;
  while (i < c.length() && c[i] == 'a') i++;
  while (i < c.length() && c[i] != 'a') {
    c[i] = char(c[i] - 1);
    i++;
  }
  cout << c;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

```