#include <bits/stdc++.h>
using namespace std;
int N, K;
void init() { scanf("%d%d", &N, &K); }
void work() {
  if (K == 1 && N > 1) {
    printf("-1");
    return;
  }
  if (K == 1 && N == 1) {
    printf("a");
    return;
  }
  if (N < K) {
    printf("-1");
    return;
  }
  if (N == K) {
    for (int i = 1; i <= K; i++) printf("%c", 'a' + i - 1);
    return;
  }
  for (int i = 1; i <= N - K + 2; i++) {
    if (i & 1)
      printf("a");
    else
      printf("b");
  }
  char j = 'c';
  for (int i = N - K + 3; i <= N; i++) printf("%c", j++);
}
int main() {
  init();
  work();
  printf("\n");
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
int N, K;
void init() { scanf("%d%d", &N, &K); }
void work() {
  if (K == 1 && N > 1) {
    printf("-1");
    return;
  }
  if (K == 1 && N == 1) {
    printf("a");
    return;
  }
  if (N < K) {
    printf("-1");
    return;
  }
  if (N == K) {
    for (int i = 1; i <= K; i++) printf("%c", 'a' + i - 1);
    return;
  }
  for (int i = 1; i <= N - K + 2; i++) {
    if (i & 1)
      printf("a");
    else
      printf("b");
  }
  char j = 'c';
  for (int i = N - K + 3; i <= N; i++) printf("%c", j++);
}
int main() {
  init();
  work();
  printf("\n");
  return 0;
}

```