#include <bits/stdc++.h>
using namespace std;
pair<int, int> operator+(const pair<int, int> &a, const pair<int, int> &b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
pair<int, int> operator-(const pair<int, int> &a, const pair<int, int> &b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
pair<int, int> &operator+=(pair<int, int> &a, const pair<int, int> &b) {
  a.first += b.first;
  a.second += b.second;
  return a;
}
pair<int, int> &operator-=(pair<int, int> &a, const pair<int, int> &b) {
  a.first -= b.first;
  a.second -= b.second;
  return a;
}
template <class T, class U>
bool cmp_second(const pair<T, U> &a, const pair<T, U> &b) {
  return a.second < b.second;
}
template <class T>
T gcd(T a, T b) {
  a = abs(a);
  b = abs(b);
  while (b) {
    T t = b;
    b = a % b;
    a = t;
  }
  return a;
}
template <class T>
pair<T, T> ext_gcd(T a, T b) {
  T a0 = 1, a1 = 0, b0 = 0, b1 = 1;
  if (a < 0) {
    a = -a;
    a0 = -1;
  }
  if (b < 0) {
    b = -b;
    b1 = -1;
  }
  while (b) {
    T t, q = a / b;
    t = b;
    b = a - b * q;
    a = t;
    t = b0;
    b0 = a0 - b0 * q;
    a0 = t;
    t = b1;
    b1 = a1 - b1 * q;
    a1 = t;
  }
  return make_pair(a0, a1);
}
inline int sg(int x) { return x ? (x > 0 ? 1 : -1) : 0; }
int n, k;
char s[1000500];
bool solve() {
  if (k > n) return 0;
  s[n] = '\0';
  if (k == 1) {
    if (n > 1)
      return 0;
    else
      s[0] = 'a';
  } else if (k == 2) {
    for (int i = 0; i < n; i++) s[i] = (i & 1) ? 'b' : 'a';
  } else {
    for (int i = 0; i < n; i++) s[i] = (i & 1) ? 'b' : 'a';
    for (int i = 2; i < k; i++) s[n - k + i] = 'a' + i;
  }
  puts(s);
  return 1;
}
int main(void) {
  scanf("%d %d", &(n), &(k));
  if (!solve()) puts("-1");
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
pair<int, int> operator+(const pair<int, int> &a, const pair<int, int> &b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
pair<int, int> operator-(const pair<int, int> &a, const pair<int, int> &b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
pair<int, int> &operator+=(pair<int, int> &a, const pair<int, int> &b) {
  a.first += b.first;
  a.second += b.second;
  return a;
}
pair<int, int> &operator-=(pair<int, int> &a, const pair<int, int> &b) {
  a.first -= b.first;
  a.second -= b.second;
  return a;
}
template <class T, class U>
bool cmp_second(const pair<T, U> &a, const pair<T, U> &b) {
  return a.second < b.second;
}
template <class T>
T gcd(T a, T b) {
  a = abs(a);
  b = abs(b);
  while (b) {
    T t = b;
    b = a % b;
    a = t;
  }
  return a;
}
template <class T>
pair<T, T> ext_gcd(T a, T b) {
  T a0 = 1, a1 = 0, b0 = 0, b1 = 1;
  if (a < 0) {
    a = -a;
    a0 = -1;
  }
  if (b < 0) {
    b = -b;
    b1 = -1;
  }
  while (b) {
    T t, q = a / b;
    t = b;
    b = a - b * q;
    a = t;
    t = b0;
    b0 = a0 - b0 * q;
    a0 = t;
    t = b1;
    b1 = a1 - b1 * q;
    a1 = t;
  }
  return make_pair(a0, a1);
}
inline int sg(int x) { return x ? (x > 0 ? 1 : -1) : 0; }
int n, k;
char s[1000500];
bool solve() {
  if (k > n) return 0;
  s[n] = '\0';
  if (k == 1) {
    if (n > 1)
      return 0;
    else
      s[0] = 'a';
  } else if (k == 2) {
    for (int i = 0; i < n; i++) s[i] = (i & 1) ? 'b' : 'a';
  } else {
    for (int i = 0; i < n; i++) s[i] = (i & 1) ? 'b' : 'a';
    for (int i = 2; i < k; i++) s[n - k + i] = 'a' + i;
  }
  puts(s);
  return 1;
}
int main(void) {
  scanf("%d %d", &(n), &(k));
  if (!solve()) puts("-1");
  return 0;
}

```