#include <bits/stdc++.h>
using namespace std;
template <class T>
inline pair<T, T> extendedEuclid(T a, T b) {
  if (b == 0)
    return pair<T, T>(1, 0);
  else {
    pair<T, T> d = extendedEuclid(b, a % b);
    return pair<T, T>(d.second, d.first - d.second * (a / b));
  }
}
template <class T>
inline T modularInverse(T a, T M) {
  pair<T, T> ret = extendedEuclid(a, M);
  return ((ret.first % M) + M) % M;
}
template <class T>
inline T bigMod(T p, T e, T M) {
  long long int ret = 1;
  while (e > 0) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
    e >>= 1;
  }
  return (T)ret;
}
template <class T>
inline T power(T p, T e) {
  long long int ret = 1;
  while (e > 0) {
    if (e & 1) ret = (ret * p);
    p = (p * p);
    e >>= 1;
  }
  return (T)ret;
}
int main(void) {
  string s, ans = ("");
  cin >> s;
  int ok = 0, i, c = 0;
  for (long long int i = (0); i < (s.length()); i++) {
    if (s[i] == 'a') c++;
  }
  if (c == s.length()) {
    s[s.length() - 1] = 'z';
    cout << s << '\n';
    return 0;
  }
  for (i = 0; i < s.length(); i++) {
    if (s[i] == 'a')
      ans += s[i];
    else
      break;
  }
  for (i; i < s.length(); i++) {
    if (s[i] == 'a')
      break;
    else
      ans += (s[i] - 1);
  }
  for (i; i < s.length(); i++) {
    ans += s[i];
  }
  cout << ans << '\n';
  return 0;
}

### 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;
template <class T>
inline pair<T, T> extendedEuclid(T a, T b) {
  if (b == 0)
    return pair<T, T>(1, 0);
  else {
    pair<T, T> d = extendedEuclid(b, a % b);
    return pair<T, T>(d.second, d.first - d.second * (a / b));
  }
}
template <class T>
inline T modularInverse(T a, T M) {
  pair<T, T> ret = extendedEuclid(a, M);
  return ((ret.first % M) + M) % M;
}
template <class T>
inline T bigMod(T p, T e, T M) {
  long long int ret = 1;
  while (e > 0) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
    e >>= 1;
  }
  return (T)ret;
}
template <class T>
inline T power(T p, T e) {
  long long int ret = 1;
  while (e > 0) {
    if (e & 1) ret = (ret * p);
    p = (p * p);
    e >>= 1;
  }
  return (T)ret;
}
int main(void) {
  string s, ans = ("");
  cin >> s;
  int ok = 0, i, c = 0;
  for (long long int i = (0); i < (s.length()); i++) {
    if (s[i] == 'a') c++;
  }
  if (c == s.length()) {
    s[s.length() - 1] = 'z';
    cout << s << '\n';
    return 0;
  }
  for (i = 0; i < s.length(); i++) {
    if (s[i] == 'a')
      ans += s[i];
    else
      break;
  }
  for (i; i < s.length(); i++) {
    if (s[i] == 'a')
      break;
    else
      ans += (s[i] - 1);
  }
  for (i; i < s.length(); i++) {
    ans += s[i];
  }
  cout << ans << '\n';
  return 0;
}

```