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;
const long long linf = (1LL << 62) - 1;
const int inf = 1010101043;
void pause(bool a = true) {}
long double PI = acos(-1);
template <class T>
inline const T sqr(const T& x) {
  return x * x;
}
template <class T>
inline T dist(pair<T, T> a, pair<T, T> b) {
  return sqrt(sqr(a.first - b.first) + sqr(a.second - b.second));
}
template <class T>
inline T prod(pair<T, T> a, pair<T, T> b, pair<T, T> c = make_pair(0, 0)) {
  return (a.first - c.first) * (b.second - c.second) -
         (a.second - c.second) * (b.first - c.first);
}
template <class T>
inline T prodv(pair<T, T> a, pair<T, T> b, pair<T, T> c = make_pair(0, 0)) {
  return (a.first - c.first) * (b.first - c.first) +
         (a.second - c.second) * (b.second - c.second);
}
template <class T>
inline bool intriangle(pair<T, T> a, pair<T, T> b, pair<T, T> c, pair<T, T> d) {
  return (abs(prod(a, b, c)) ==
          abs(prod(a, b, d)) + abs(prod(a, c, d)) + abs(prod(b, c, d)));
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long popcnt(long long mask) {
  long long cnt = 0;
  while (mask) cnt += mask & 1, mask >>= 1;
  return cnt;
}
unsigned long long powi(unsigned long long a, unsigned long long b,
                        long long mod = 1000000007) {
  if (b < 2) return b ? a : 1;
  return (powi((a * a) % mod, b / 2, mod) * (b % 2 ? a : 1LL)) % mod;
}
template <class T>
bool maxz(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool minz(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool maxze(T& a, T b) {
  if (a <= b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool minze(T& a, T b) {
  if (a >= b) {
    a = b;
    return true;
  }
  return false;
}
long long mod7 = 1000000007;
long long mod9 = 1000000009;
bool isprime(long long n) {
  if (n < 2) return 0;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
long long get(long long x, vector<long long>& a) {
  long long d = 0;
  for (int i = 0; i < (a.size()); i++) d += abs(a[i] - x);
  return d;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  long long n = s.length();
  bool st = false;
  for (int i = 0; i < (n); i++) {
    if (s[i] == 'a') {
      if (st) break;
      continue;
    } else {
      st = true;
      s[i] = s[i] - 1;
    }
  }
  if (st == false) s[n - 1] = 'z';
  cout << s << endl;
  pause(0);
  return 0;
}
