#include <bits/stdc++.h>
int main() {
  int n, k, tmp;
  char ac;
  while (scanf("%d%d", &n, &k) == 2) {
    if (n < k) {
      printf("-1\n");
      continue;
    }
    if (n > 1 && k == 1) {
      printf("-1\n");
      continue;
    }
    if (n == 1 && k == 1) {
      printf("a\n");
      continue;
    }
    tmp = n - k + 2;
    for (int i = 0; i < tmp; i++) {
      if (i & 1)
        printf("b");
      else
        printf("a");
    }
    tmp = k - 2;
    ac = 'c';
    for (int i = 0; i < tmp; i++) {
      printf("%c", ac);
      ac++;
    }
    printf("\n");
  }
}

### Prompt

Generate a CPP solution to 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>
int main() {
  int n, k, tmp;
  char ac;
  while (scanf("%d%d", &n, &k) == 2) {
    if (n < k) {
      printf("-1\n");
      continue;
    }
    if (n > 1 && k == 1) {
      printf("-1\n");
      continue;
    }
    if (n == 1 && k == 1) {
      printf("a\n");
      continue;
    }
    tmp = n - k + 2;
    for (int i = 0; i < tmp; i++) {
      if (i & 1)
        printf("b");
      else
        printf("a");
    }
    tmp = k - 2;
    ac = 'c';
    for (int i = 0; i < tmp; i++) {
      printf("%c", ac);
      ac++;
    }
    printf("\n");
  }
}

```