Problem:
Score : 400 points

Problem Statement

Mr. Takahashi has a string s consisting of lowercase English letters.
He repeats the following operation on s exactly K times.

Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of z is a.

For example, if you perform an operation for the second letter on aaz, aaz becomes abz.
If you then perform an operation for the third letter on abz, abz becomes aba.

Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s.
Find the such string.

Constraints

1≤|s|≤10^5

All letters in s are lowercase English letters.

1≤K≤10^9

Input

The input is given from Standard Input in the following format:

s
K

Output

Print the lexicographically smallest string after performing exactly K operations on s.

Sample Input 1

xyz
4

Sample Output 1

aya

For example, you can perform the following operations: xyz, yyz, zyz, ayz, aya.

Sample Input 2

a
25

Sample Output 2

z

You have to perform exactly K operations.

Sample Input 3

codefestival
100

Sample Output 3

aaaafeaaivap

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

// #define fi first
// #define se second
#define rep(i, n) for (ll i=0;i<(n); i++)
#define rrep(i, n) for (ll (i) = n-1;(i) >= 0;--(i))
#define ALL(v) (v).begin(), (v).end()
#define FORR(i, a, b) for(int i=(b)-1; i>=(a); --i)
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pi;

//const ll M=1000000007ll;
//const ll MOD=1000000007ll;
const ll MOD=2019ll;
const ll MAX=100000001ll;
//const ll M = 1000000000+7;

//const int mod = 1000000007;
const int mod = 2019;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
   return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
 
  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

mint f(ll n){
  if(n==0) return 1;
  mint ans = f(n/2);
  ans *= ans;
  if(n%2==1) ans*= 2;
  return ans;
}

mint g(ll n, ll x){
  if(x==0) return 1;
  mint ans = g(n, x/2);
  ans *= ans;
  if(x%2==1) ans*= n;
  return ans;  
}

mint choose(ll n, ll k){
  mint x=1, y=1;
  rep(i,k){
    x *= (n-i);
    y *= (i+1);
  }
  return x / y;
}

struct BIT {
  private:
   vector<ll> bit;
   ll N;
 
  public:
   BIT(ll size) {
     N = size;
     bit.resize(N + 1);
   }
 
   // 一点更新です
   void add(ll a, ll w) {
     for (ll x = a; x <= N; x += x & -x) bit[x] += w;
   }
 
   // 1~Nまでの和を求める。
   int sum(ll a) {
     ll ret = 0;
     for (ll x = a; x > 0; x -= x & -x) ret += bit[x];
     return ret;
   }
 };
 
ll fff(ll n, vector<ll>& v, ll i, ll j, ll rl){
  ll can = 0;
  ll can_i = 0;

  cout << v[0]<<endl;
  
  // right
  if(rl == 0){
    rep(k, n){
      if(v[k]*(j-k)>can){
	can = v[k]*(j-k);
	can_i = k;
      }
      //cout << can<<endl;
    }
    v[can_i] = 0;
    return can;
  }
  // left
  else{
    rep(k, n){
      if(v[k]*(k-i)>can){
	can = v[k]*(k-i);
	can_i = k;
      }
      //cout << can<<endl;

    }
    v[can_i] = 0;
    return can;
  }
  
}

ll ff(ll x){
  return x * (x+1) / 2;
}

void dfs(string s, ll ma, ll n){
  //cout << ma<<endl;
  if(s.size()==n){
    cout << s <<endl;
    return;
  }
  rep(i, ma+1){
    dfs(s + char('a'+i), max(ma, i+1), n);
  }
}

int main(){
  string s;
  cin>>s;
  ll k;
  cin>>k;

  ll keta = 0;
  while(keta<s.size()){
    if(k>=(26-(s[keta]-'a'))){
      k -= 26-(s[keta] - 'a');
      s[keta] = 'a';
    }
    keta++;
  }
  ll tmp = (s[s.size()-1]-'a');
  tmp = (tmp+k)%26;
  s[s.size()-1] = char(tmp+'a');
  cout << s << endl;
}