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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <fstream>
#include <vector>
#include <deque>
#include <assert.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <stdio.h>
#include <string.h>
#include <utility>
#include <math.h>
#include <bitset>
#include <iomanip>
#include <complex>
using namespace std;
#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define per(i,b) gnr(i,0,b)
#define ALL(x) (x).begin(),(x).end()
#define RALL(x) (x).rbegin(),(x).rend()
#define IDX(vec, element_iter) distance((vec).begin(), element_iter)
#define print(x) cout<<(x)<<'\n'
#define print_vec(v) rep(i,v.size()){if(i!=0)cout<<" ";cout<<v[i];}cout<<'\n'
#define pe(x) cout<<(x)<<" "
#define DEBUG(x) cout<<#x<<": "<<x<<'\n'
#define DEBUG_VEC(v) cout<<#v<<":";rep(i,v.size())cout<<" "<<v[i];cout<<'\n'
#define pb push_back
#define mp make_pair
#define MOD 1000000007 //10^9+7
#define fi first
#define sc second
using ll=long long;
using pi=pair<int,int>;
using pll=pair<ll,ll>;
using vi=vector<int>;
using vll=vector<ll>;
using vstr=vector<string>;
int dx[4]={0,0,-1,1};
int dy[4]={-1,1,0,0};
void Yes(bool c) { if (c)print("Yes"); else print("No"); };
void YES(bool c) { if (c)print("YES"); else print("NO"); };

void solve() {
    string s; cin >> s;
    int length = s.size();
    string ans = "";
    ll K; cin >> K;
    rep(i,length){
        char t;
        int ascii = (int) s[i];
        int diff = ascii-97;
        int x = 26 - diff;
        if(diff == 0) {
            t = s[i];
        }else if(x > K){
            t = s[i];
        }else{
            t = 'a';
            K -= x;
        }
        if(i == length-1){
            t += K%26;
        }
        ans = ans + t;
    }
    print(ans);
  
}
 
int main() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
	cout<<fixed<<setprecision(20);
    //int q; cin >> q;
	//while (q--)
		solve();
    return 0;
}