#include <bits/stdc++.h>
using namespace std;
template <class T>
string str(T Number) {
  string Result;
  ostringstream convert;
  convert << Number;
  Result = convert.str();
  return Result;
}
int StringToNumber(const string &Text) {
  istringstream ss(Text);
  int result;
  return ss >> result ? result : 0;
}
template <class T>
inline vector<pair<T, int> > FACTORISE(T n) {
  vector<pair<T, int> > R;
  for (T i = 2; n > 1;) {
    if (n % i == 0) {
      int C = 0;
      for (; n % i == 0; C++, n /= i)
        ;
      R.push_back(make_pair(i, C));
    }
    i++;
    if (i > n / i) i = n;
  }
  if (n > 1) R.push_back(make_pair(n, 1));
  return R;
}
template <class T>
inline T TOTIENT(T n) {
  vector<pair<T, int> > R = FACTORISE(n);
  T r = n;
  for (int i = 0; i < R.size(); i++) r = r / R[i].first * (R[i].first - 1);
  return r;
}
double rnd(float d) { return floor(d + 0.49); }
long long power(long long a, long long b, long long mod) {
  long long ret = 1;
  while (b > 0) {
    if (b & 1) ret = (ret * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return ret % mod;
}
int main() {
  int n, k;
  cin >> n >> k;
  if (n < k)
    cout << -1 << endl;
  else {
    if (n == 1)
      cout << "a" << endl;
    else if (n == 2) {
      if (k == 2)
        cout << "ab" << endl;
      else
        cout << "aa" << endl;
    } else {
      if (k == 1) {
        cout << -1 << endl;
        return 0;
      } else {
        string y = "";
        int co = 0;
        for (int i = 0; i < n - k + 2; i++) {
          if (co == 0)
            y += "a";
          else
            y += "b";
          co = (co + 1) % 2;
        }
        int h = 1;
        for (int i = n - k + 2; i < n; i++) {
          y += 'b' + h;
          h++;
        }
        cout << y << endl;
      }
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

Little penguin Polo adores strings. But most of all he adores strings of length n.

One day he wanted to find a string that meets the following conditions:

  1. The string consists of n lowercase English letters (that is, the string's length equals n), exactly k of these letters are distinct. 
  2. No two neighbouring letters of a string coincide; that is, if we represent a string as s = s1s2... sn, then the following inequality holds, si ≠ si + 1(1 ≤ i < n). 
  3. Among all strings that meet points 1 and 2, the required string is lexicographically smallest. 



Help him find such string or state that such string doesn't exist.

String x = x1x2... xp is lexicographically less than string y = y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or there is such number r (r < p, r < q), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1. The characters of the strings are compared by their ASCII codes.

Input

A single line contains two positive integers n and k (1 ≤ n ≤ 106, 1 ≤ k ≤ 26) — the string's length and the number of distinct letters.

Output

In a single line print the required string. If there isn't such string, print "-1" (without the quotes).

Examples

Input

7 4


Output

ababacd


Input

4 7


Output

-1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
string str(T Number) {
  string Result;
  ostringstream convert;
  convert << Number;
  Result = convert.str();
  return Result;
}
int StringToNumber(const string &Text) {
  istringstream ss(Text);
  int result;
  return ss >> result ? result : 0;
}
template <class T>
inline vector<pair<T, int> > FACTORISE(T n) {
  vector<pair<T, int> > R;
  for (T i = 2; n > 1;) {
    if (n % i == 0) {
      int C = 0;
      for (; n % i == 0; C++, n /= i)
        ;
      R.push_back(make_pair(i, C));
    }
    i++;
    if (i > n / i) i = n;
  }
  if (n > 1) R.push_back(make_pair(n, 1));
  return R;
}
template <class T>
inline T TOTIENT(T n) {
  vector<pair<T, int> > R = FACTORISE(n);
  T r = n;
  for (int i = 0; i < R.size(); i++) r = r / R[i].first * (R[i].first - 1);
  return r;
}
double rnd(float d) { return floor(d + 0.49); }
long long power(long long a, long long b, long long mod) {
  long long ret = 1;
  while (b > 0) {
    if (b & 1) ret = (ret * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return ret % mod;
}
int main() {
  int n, k;
  cin >> n >> k;
  if (n < k)
    cout << -1 << endl;
  else {
    if (n == 1)
      cout << "a" << endl;
    else if (n == 2) {
      if (k == 2)
        cout << "ab" << endl;
      else
        cout << "aa" << endl;
    } else {
      if (k == 1) {
        cout << -1 << endl;
        return 0;
      } else {
        string y = "";
        int co = 0;
        for (int i = 0; i < n - k + 2; i++) {
          if (co == 0)
            y += "a";
          else
            y += "b";
          co = (co + 1) % 2;
        }
        int h = 1;
        for (int i = n - k + 2; i < n; i++) {
          y += 'b' + h;
          h++;
        }
        cout << y << endl;
      }
    }
  }
  return 0;
}

```