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 <algorithm>
#include <iostream>
#include <vector>
using namespace std;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}

typedef long long ll;
ll MOD = 1000000007;

// a^p
// 2^3 = 2 * 2^2
// 2^2 = 2 * (2^1)
// 2^1 = 2
ll modpow(ll a, ll p, ll mod) {
  if (p == 0)
    return 1;

  if (p % 2 == 0) {
    ll half = modpow(a, p / 2, mod) % mod;
    return half * half % mod;
  } else {
    return a * modpow(a, p - 1, mod) % mod;
  }
}

// nCa を求める
ll modCombination(ll n, ll a, ll mod) {
  if (n - a < a) {
    return modCombination(n, n - a, mod);
  }

  ll denominator = 1; // 分母
  ll numerator = 1;   // 分子

  for (ll i = 0; i < a; i++) {
    denominator *= a - i;
    numerator *= n - i;
    denominator %= mod;
    numerator %= mod;
  }

  return numerator * modpow(denominator, mod - 2, mod) % mod;
}

class UnionFind {
private:
  vector<ll> parents;

public:
  UnionFind(ll n) : parents(n, -1) {}

  bool issame(ll x, ll y) {
    return root(x) == root(y);
  }

  bool merge(ll x, ll y) {
    if (issame(x, y))
      return false;

    ll rx = root(x);
    ll ry = root(y);
    if (parents[rx] > parents[ry])
      swap(rx, ry);
    // サイズ情報を更新
    parents[rx] += parents[ry];
    // yの親を更新
    parents[ry] = rx;

    return true;
  }

  ll size(ll x) {
    return -parents[root(x)];
  }

private:
  ll root(ll x) {
    if (parents[x] < 0)
      return x;
    // 根の親の値に木の(-)サイズの情報を入れる
    return parents[x] = root(parents[x]);
  }
};

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

  ll remain = K;
  string ans = s;
  for (ll i = 0; i < s.size(); i++) {
    const char c = s[i];
    ll n = (ll)('z' - c + 1) % 26;

    if (remain >= n) {
      remain -= n;
      ans[i] = 'a';
    }

    if (i == s.size() - 1 && remain > 0) {
      ans[i] = (char)('a' + (c + remain - 'a') % 26);
    }
  }

  cout << ans << endl;
  return 0;
}