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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<string>
using namespace std;
int gcd(int a, int b) {
	int c = a % b;
	while (c != 0) {
		a = b;
		b = c;
		c = a % b;
	}
	return b;
}
struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++) par[i] = i;
	}

	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}

	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};
typedef long long ll;

ll M = 1000000007;

vector<ll> fac(300001); //n!(mod M)
vector<ll> ifac(300001); //k!^{M-2} (mod M)

ll mpow(ll x, ll n) {
	ll ans = 1;
	while (n != 0) {
		if (n & 1) ans = ans * x % M;
		x = x * x % M;
		n = n >> 1;
	}
	return ans;
}
void setcomb() {
	fac[0] = 1;
	ifac[0] = 1;
	for (ll i = 0; i < 300000; i++) {
		fac[i + 1] = fac[i] * (i + 1) % M; // n!(mod M)
		ifac[i + 1] = ifac[i] * mpow(i + 1, M - 2) % M; // k!^{M-2} (mod M) ←累乗にmpowを採用
	}
}
ll comb(ll a, ll b) {
	if (a == 0 && b == 0)return 1;
	if (a < b || a < 0)return 0;
	ll tmp = ifac[a - b] * ifac[b] % M;
	return tmp * fac[a] % M;
}
// mod. m での a の逆元 a^{-1} を計算する
long long modinv(long long a) {
	long long b = M, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= M;
	if (u < 0) u += M;
	return u;
}
vector<vector<ll>> mul(vector<vector<ll>> a, vector<vector<ll>> b, int n) {
	int i, j, k, t;
	vector<vector<ll>> c(n);
	for (i = 0; i < n; i++) {
		for (j = 0; j < n; j++) {
			t = 0;
			for (k = 0; k < n; k++)
				t = (t + a[i][k] * b[k][j] % M) % M;
			c[i].push_back(t);
		}
	}
	return c;
}

template< typename Monoid >
struct SegmentTree {

	int sz;
	vector< Monoid > seg;

	const Monoid M1;

	SegmentTree(int n, const Monoid &M1) : M1(M1) {
		sz = 1;
		while (sz < n) sz <<= 1;
		seg.assign(2 * sz, M1);
	}

	void set(int k, const Monoid &x) {
		seg[k + sz] = x;
	}

	void build() {
		for (int k = sz - 1; k > 0; k--) {
			seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
		}
	}

	void update(int k, const Monoid &x) {
		k += sz;
		seg[k] = x;
		while (k >>= 1) {
			seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
		}
	}

	Monoid query(int a, int b) {
		Monoid L = M1, R = M1;
		for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
			if (a & 1) L = f(L, seg[a++]);
			if (b & 1) R = f(seg[--b], R);
		}
		return f(L, R);
	}

	Monoid value(const int &k) const {
		return seg[k + sz];
	}

	Monoid f(Monoid a, Monoid b) {
		return a + b;
	}
};

int main() {
	int i, k;
	char s[111111];
	cin >> s >> k;
	for (i = 0; s[i] != '\0'; i++) {
		if (s[i] != 'a') {
			if (k > 'z' - s[i]) {
				k -= 'z' - s[i] + 1;
				s[i] = 'a';
			}
		}
	}
	cout << s << endl;
}