#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (a < b)
    return gcd(b, a);
  else if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int isPrime(long long int n) {
  long long int p = (long long int)sqrt(n);
  for (long long int i = (2); i <= (p); i += (1))
    if (n % i == 0) return 0;
  return 1;
}
long long int pow(long long int b, long long int e) {
  if (e == 0)
    return 1;
  else if (e % 2 == 0) {
    long long int a = pow(b, e / 2);
    return a * a;
  } else {
    long long int a = pow(b, e / 2);
    return b * a * a;
  }
}
long long int powm(long long int x, long long int y,
                   long long int m = 1000000007) {
  x = x % m;
  long long int res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int modInverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
bool issq(long long int n) {
  long long int p = sqrt(n);
  return p * p == n;
}
vector<long long int> prime;
void seive(long long int n) {
  prime.resize(n + 1);
  for (long long int i = (2); i <= (n); i += (1))
    if (prime[i] == 0) {
      prime[i] = i;
      for (long long int j = (i * i); j <= (n); j += (i))
        if (prime[j] == 0) prime[j] = i;
    }
}
long long int ext_gcd(long long int A, long long int B, long long int *X,
                      long long int *Y) {
  long long int x2, y2, x1, y1, x, y, r2, r1, q, r;
  x2 = 1;
  y2 = 0;
  x1 = 0;
  y1 = 1;
  for (r2 = A, r1 = B; r1 != 0;
       r2 = r1, r1 = r, x2 = x1, y2 = y1, x1 = x, y1 = y) {
    q = r2 / r1;
    r = r2 % r1;
    x = x2 - (q * x1);
    y = y2 - (q * y1);
  }
  *X = x2;
  *Y = y2;
  return r2;
}
long long int mulInv(long long int a, long long int m) {
  long long int x, y;
  ext_gcd(a, m, &x, &y);
  x %= m;
  if (x < 0) x += m;
  return x;
}
long long int find(long long int num[], long long int rem[], long long int k,
                   long long int prod) {
  long long int result = 0;
  for (long long int i = 0; i < k; i++) {
    long long int pp = prod / num[i];
    result += rem[i] * mulInv(pp, num[i]) * pp;
  }
  return result % prod;
}
class DSU {
 public:
  vector<long long int> a, rank;

 public:
  long long int c;
  map<string, long long int> m;

 public:
  DSU(long long int n) {
    c = 1;
    a.resize(n + 1);
    rank.resize(n + 1);
    for (long long int i = 1; i <= n; i++) a[i] = i;
  }

 public:
  long long int root(long long int x) {
    long long int u = x;
    while (a[x] != x) {
      a[x] = a[a[x]];
      x = a[x];
    }
    a[u] = x;
    return x;
  }

 public:
  void unite(long long int x, long long int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (rank[x] < rank[y])
        a[y] = x;
      else
        a[x] = y;
    }
  }
};
struct trip {
  long long int l1, l2, r2, r1, ind;
};
bool comp(trip p1, trip p2) {
  return (p1.l2 > p2.l2 && (p1.l2 == p2.l2 && p1.r2 > p2.r2));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  string s;
  cin >> n >> k >> s;
  vector<char> v(n);
  for (long long int i = (0); i <= (n - 1); i += (1)) {
    long long int c = s[i] - 96;
    long long int d = max(26 - c, c - 1);
    d = min(k, d);
    if (d < c)
      v[i] = s[i] - d;
    else
      v[i] = s[i] + d;
    k -= d;
  }
  if (k == 0)
    for (auto x : v) cout << x;
  else
    cout << -1;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

Limak is a little polar bear. He likes nice strings — strings of length n, consisting of lowercase English letters only.

The distance between two letters is defined as the difference between their positions in the alphabet. For example, <image>, and <image>.

Also, the distance between two nice strings is defined as the sum of distances of corresponding letters. For example, <image>, and <image>.

Limak gives you a nice string s and an integer k. He challenges you to find any nice string s' that <image>. Find any s' satisfying the given conditions, or print "-1" if it's impossible to do so.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 106).

The second line contains a string s of length n, consisting of lowercase English letters.

Output

If there is no string satisfying the given conditions then print "-1" (without the quotes).

Otherwise, print any nice string s' that <image>.

Examples

Input

4 26
bear


Output

roar

Input

2 7
af


Output

db


Input

3 1000
hey


Output

-1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (a < b)
    return gcd(b, a);
  else if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int isPrime(long long int n) {
  long long int p = (long long int)sqrt(n);
  for (long long int i = (2); i <= (p); i += (1))
    if (n % i == 0) return 0;
  return 1;
}
long long int pow(long long int b, long long int e) {
  if (e == 0)
    return 1;
  else if (e % 2 == 0) {
    long long int a = pow(b, e / 2);
    return a * a;
  } else {
    long long int a = pow(b, e / 2);
    return b * a * a;
  }
}
long long int powm(long long int x, long long int y,
                   long long int m = 1000000007) {
  x = x % m;
  long long int res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int modInverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
bool issq(long long int n) {
  long long int p = sqrt(n);
  return p * p == n;
}
vector<long long int> prime;
void seive(long long int n) {
  prime.resize(n + 1);
  for (long long int i = (2); i <= (n); i += (1))
    if (prime[i] == 0) {
      prime[i] = i;
      for (long long int j = (i * i); j <= (n); j += (i))
        if (prime[j] == 0) prime[j] = i;
    }
}
long long int ext_gcd(long long int A, long long int B, long long int *X,
                      long long int *Y) {
  long long int x2, y2, x1, y1, x, y, r2, r1, q, r;
  x2 = 1;
  y2 = 0;
  x1 = 0;
  y1 = 1;
  for (r2 = A, r1 = B; r1 != 0;
       r2 = r1, r1 = r, x2 = x1, y2 = y1, x1 = x, y1 = y) {
    q = r2 / r1;
    r = r2 % r1;
    x = x2 - (q * x1);
    y = y2 - (q * y1);
  }
  *X = x2;
  *Y = y2;
  return r2;
}
long long int mulInv(long long int a, long long int m) {
  long long int x, y;
  ext_gcd(a, m, &x, &y);
  x %= m;
  if (x < 0) x += m;
  return x;
}
long long int find(long long int num[], long long int rem[], long long int k,
                   long long int prod) {
  long long int result = 0;
  for (long long int i = 0; i < k; i++) {
    long long int pp = prod / num[i];
    result += rem[i] * mulInv(pp, num[i]) * pp;
  }
  return result % prod;
}
class DSU {
 public:
  vector<long long int> a, rank;

 public:
  long long int c;
  map<string, long long int> m;

 public:
  DSU(long long int n) {
    c = 1;
    a.resize(n + 1);
    rank.resize(n + 1);
    for (long long int i = 1; i <= n; i++) a[i] = i;
  }

 public:
  long long int root(long long int x) {
    long long int u = x;
    while (a[x] != x) {
      a[x] = a[a[x]];
      x = a[x];
    }
    a[u] = x;
    return x;
  }

 public:
  void unite(long long int x, long long int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (rank[x] < rank[y])
        a[y] = x;
      else
        a[x] = y;
    }
  }
};
struct trip {
  long long int l1, l2, r2, r1, ind;
};
bool comp(trip p1, trip p2) {
  return (p1.l2 > p2.l2 && (p1.l2 == p2.l2 && p1.r2 > p2.r2));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  string s;
  cin >> n >> k >> s;
  vector<char> v(n);
  for (long long int i = (0); i <= (n - 1); i += (1)) {
    long long int c = s[i] - 96;
    long long int d = max(26 - c, c - 1);
    d = min(k, d);
    if (d < c)
      v[i] = s[i] - d;
    else
      v[i] = s[i] + d;
    k -= d;
  }
  if (k == 0)
    for (auto x : v) cout << x;
  else
    cout << -1;
  return 0;
}

```