#include <bits/stdc++.h>
using namespace std;
const int INF = 1000000002;
const long long INFLL = 100000000000000000ll;
const long long MOD = 1000000007;
int N, K;
string S;
int main() {
  cin >> N >> K;
  if (K == 1) {
    if (N > 1)
      cout << -1;
    else
      cout << "a";
    return 0;
  }
  if (K > N) {
    cout << -1;
    return 0;
  }
  for (int i = (1); i <= (N - K + 2); i++) {
    if (i % 2)
      S += 'a';
    else
      S += 'b';
  }
  for (int i = (N - K + 3); i <= (N); i++) S += (char)(i - (N - K + 1) + 'a');
  cout << S;
}

### Prompt

Please create a solution in cpp 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;
const int INF = 1000000002;
const long long INFLL = 100000000000000000ll;
const long long MOD = 1000000007;
int N, K;
string S;
int main() {
  cin >> N >> K;
  if (K == 1) {
    if (N > 1)
      cout << -1;
    else
      cout << "a";
    return 0;
  }
  if (K > N) {
    cout << -1;
    return 0;
  }
  for (int i = (1); i <= (N - K + 2); i++) {
    if (i % 2)
      S += 'a';
    else
      S += 'b';
  }
  for (int i = (N - K + 3); i <= (N); i++) S += (char)(i - (N - K + 1) + 'a');
  cout << S;
}

```