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;
bool isPrime[1000005] = {false};
vector<long long> prm;
vector<long long> SPF(1000005);
void sieve() {
  isPrime[1] = isPrime[0] = 0;
  long long i;
  for (i = 2; i < 1000005; ++i) isPrime[i] = 1;
  for (i = 2; i * i < 1000005; ++i) {
    if (isPrime[i]) {
      for (long long j = i * i; j < 1000005; j += i) isPrime[j] = 0;
    }
  }
  if (!prm.empty()) prm.clear();
  for (i = 2; i < 1000005; ++i) {
    if (isPrime[i]) prm.push_back(i);
  }
}
bool flip(bool f) {
  if (f) return 0;
  return 1;
}
long long fastPower(long long a, long long b) {
  if (b == 0) return 1;
  long long prod = 1;
  while (b > 0) {
    if (b & 1) prod = prod * a;
    b = b >> 1;
    a = a * a;
  }
  return prod;
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) {
    res *= i;
  }
  return res;
}
long long nCr(long long n, long long r) {
  return (long double)fact(n) / (long double)(fact(r) * fact(n - r));
}
void printa(bool f) {
  if (f)
    cout << "YES"
         << "\n";
  else
    cout << "NO"
         << "\n";
}
void solve() {
  long long n, i;
  string s;
  cin >> s;
  n = s.length();
  long long cnt = 0;
  i = 0;
  for (; i < n; ++i) {
    if (s[i] != 'a') ++cnt;
    if (cnt > 0 && s[i] == 'a') break;
    if (cnt) s[i] = s[i] - 1;
  }
  if (cnt == 0) s[n - 1] = 'z';
  cout << s << "\n";
}
int main() {
  int tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}
