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.
#include <bits/stdc++.h>
using namespace std;
string a;
string b;
int main() {
  cin >> a;
  int index = -1;
  int len = a.length();
  int j;
  for (int i = 0; i < len; i++) {
    if (a[i] == 'a') {
      index = i;
      break;
    }
  }
  if (index == -1) {
    for (int i = 0; i < len; i++) {
      b.push_back(a[i] - 1);
    }
    cout << b << endl;
  } else {
    if (index == 0) {
      while (index < len && a[index] == 'a') index++;
      if (index == len) {
        for (int i = 0; i < len - 1; i++) b.push_back(a[i]);
        b.push_back('z');
        cout << b << endl;
      } else {
        j = index;
        for (; j < len; j++) {
          if (a[j] == 'a') break;
        }
        if (j == len) {
          for (int i = 0; i < index; i++) {
            b.push_back(a[i]);
          }
          j = index;
          for (; j < len; j++) {
            b.push_back(a[j] - 1);
          }
          cout << b << endl;
        } else {
          for (int i = 0; i < index; i++) {
            b.push_back(a[i]);
          }
          for (int i = index; i < j; i++) {
            b.push_back(a[i] - 1);
          }
          for (int i = j; i < len; i++) {
            b.push_back(a[i]);
          }
          cout << b << endl;
        }
      }
    } else {
      for (int i = 0; i < index; i++) {
        b.push_back(a[i] - 1);
      }
      j = index;
      for (; j < len; j++) {
        b.push_back(a[j]);
      }
      cout << b << endl;
    }
  }
}
