#include <bits/stdc++.h>
int dx[] = {0, 1, 0, -1, 1, -1, -1, 1};
int dy[] = {1, 0, -1, 0, 1, 1, -1, -1};
long long int gcd(long long int a, long long int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
using namespace std;
int main() {
  std::ios_base::sync_with_stdio(false);
  int i, j, k, it, t;
  {
    string s, a, b;
    cin >> s;
    a = s;
    b = s;
    for (i = 0; i < s.size(); i++) {
      if (a[i] == 'a') break;
      a[i] = 'a' + (a[i] - 'a' + 25) % 26;
    }
    for (i = 0; i < s.size(); i++) {
      if (s[i] == 'a') {
        while (s[i] == 'a') {
          i++;
        }
        break;
      }
    }
    for (j = i; j < s.size(); j++) {
      if (b[j] == 'a') break;
      b[j] = 'a' + (b[j] - 'a' + 25) % 26;
    }
    if (min(a, b) == s) {
      s[s.size() - 1] = 'z';
      cout << s << endl;
    } else
      cout << min(a, b) << endl;
  }
  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>
int dx[] = {0, 1, 0, -1, 1, -1, -1, 1};
int dy[] = {1, 0, -1, 0, 1, 1, -1, -1};
long long int gcd(long long int a, long long int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
using namespace std;
int main() {
  std::ios_base::sync_with_stdio(false);
  int i, j, k, it, t;
  {
    string s, a, b;
    cin >> s;
    a = s;
    b = s;
    for (i = 0; i < s.size(); i++) {
      if (a[i] == 'a') break;
      a[i] = 'a' + (a[i] - 'a' + 25) % 26;
    }
    for (i = 0; i < s.size(); i++) {
      if (s[i] == 'a') {
        while (s[i] == 'a') {
          i++;
        }
        break;
      }
    }
    for (j = i; j < s.size(); j++) {
      if (b[j] == 'a') break;
      b[j] = 'a' + (b[j] - 'a' + 25) % 26;
    }
    if (min(a, b) == s) {
      s[s.size() - 1] = 'z';
      cout << s << endl;
    } else
      cout << min(a, b) << endl;
  }
  return 0;
}

```