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;
inline long long get_num(char ch) {
  if (ch == '-')
    return 0;
  else if (ch >= 'a' && ch <= 'z') {
    return 1 + (ch - 'a');
  } else if (ch >= 'A' && ch <= 'Z') {
    return 27 + (ch - 'A');
  } else if (ch >= '0' && ch <= '9') {
    return 53 + (ch - '0');
  }
}
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
inline int GA(long long n, vector<long long>& vec) {
  for (long long i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    vec.push_back(x);
  }
  return 0;
}
inline int GAEG(long long m, vector<long long> vec[]) {
  for (long long i = 0; i < m; ++i) {
    long long x, y;
    cin >> x >> y;
    x--, y--;
    vec[x].push_back(y);
    vec[y].push_back(x);
  }
  return 0;
}
long long pw(long long x, long long y) {
  if (y == 0) return 1;
  long long o = y / 2;
  long long f = pw(x, o);
  if (y % 2) {
    return f * f * x;
  } else
    return f * f;
}
const long long MD = 1e9 + 7;
using namespace std;
int main() {
  string s;
  cin >> s;
  bool bl = false;
  for (long long i = 0; i < s.size(); ++i) {
    if (s[i] != 'a') {
      long long j = i;
      while (j < s.size() && s[j] != 'a') s[j]--, j++;
      bl = true;
      break;
    }
  }
  if (!bl) s[s.size() - 1] = 'z';
  cout << s << endl;
  return 0;
}
