#include <bits/stdc++.h>
using namespace std;
int main() {
  if (0) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  int n, m;
  scanf("%ld%ld", &n, &m);
  if (n < m | (n > 1 && m == 1))
    printf("-1");
  else if (m == 1)
    printf("a");
  else {
    for (int i = 0; i < n - m + 2; i++) printf("%c", 'a' + (i & 1));
    for (int i = 2; i < m; i++) printf("%c", 'a' + i);
  }
}

### Prompt

Please provide a Cpp coded solution 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;
int main() {
  if (0) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  int n, m;
  scanf("%ld%ld", &n, &m);
  if (n < m | (n > 1 && m == 1))
    printf("-1");
  else if (m == 1)
    printf("a");
  else {
    for (int i = 0; i < n - m + 2; i++) printf("%c", 'a' + (i & 1));
    for (int i = 2; i < m; i++) printf("%c", 'a' + i);
  }
}

```