#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline string tostring(const T &x) {
  ostringstream os;
  os << x;
  return os.str();
}
inline int toint(const string &s) {
  istringstream is(s);
  int x;
  is >> x;
  return x;
}
inline int todecimal(string s) {
  int a = 0;
  for (int i = 0; i < s.size(); i++) a = 2 * a + (s[i] - '0');
  return a;
}
inline string tobinary(int a) {
  string s;
  while (a != 0) {
    s = (char)(a % 2 + '0') + s;
    a >>= 1;
  }
  return s;
}
template <typename T>
inline T sqr(T x) {
  return x * x;
}
template <typename T>
T gcd(T a, T b) {
  return (b == 0) ? abs(a) : gcd(b, a % b);
}
inline int isvowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') return 1;
  return 0;
}
inline int isprime(int a) {
  for (int i = 2; i * i <= a; i++)
    if (!(a % i)) return 0;
  return 1;
}
inline void inp(int &n) {
  n = 0;
  int ch = getchar();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    n = (n << 3) + (n << 1) + ch - '0', ch = getchar();
  n = n * sign;
}
const int dx4[4] = {0, 0, 1, -1};
const int dy4[4] = {1, -1, 0, 0};
const int dx8[8] = {0, 0, 1, 1, 1, -1, -1, -1};
const int dy8[8] = {1, -1, 0, 1, -1, 1, 0, -1};
class node {
 public:
  int x, y, z;
  node(int a, int b, int c) {
    x = a;
    y = b;
    z = c;
  }
};
bool operator<(const node &a, const node &b) { return a.x > b.x; }
bool compare(const node &a, const node &b) { return a.x < b.x; }
int n, k;
void solve() {
  if (k > n || k == 1 && n > 1)
    cout << -1 << endl;
  else {
    if (n == 1 && k == 1)
      cout << "a\n";
    else {
      for (int i = 1; i <= n - k + 2; i++) {
        if (i & 1)
          cout << "a";
        else
          cout << "b";
      }
      for (int i = 1; i <= k - 2; i++) {
        cout << char('b' + i);
      }
      cout << endl;
    }
  }
}
void input() { cin >> n >> k; }
int main() {
  int testcase = 1;
  for (int i = 0; i < testcase; i++) {
    input();
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
template <typename T>
inline string tostring(const T &x) {
  ostringstream os;
  os << x;
  return os.str();
}
inline int toint(const string &s) {
  istringstream is(s);
  int x;
  is >> x;
  return x;
}
inline int todecimal(string s) {
  int a = 0;
  for (int i = 0; i < s.size(); i++) a = 2 * a + (s[i] - '0');
  return a;
}
inline string tobinary(int a) {
  string s;
  while (a != 0) {
    s = (char)(a % 2 + '0') + s;
    a >>= 1;
  }
  return s;
}
template <typename T>
inline T sqr(T x) {
  return x * x;
}
template <typename T>
T gcd(T a, T b) {
  return (b == 0) ? abs(a) : gcd(b, a % b);
}
inline int isvowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') return 1;
  return 0;
}
inline int isprime(int a) {
  for (int i = 2; i * i <= a; i++)
    if (!(a % i)) return 0;
  return 1;
}
inline void inp(int &n) {
  n = 0;
  int ch = getchar();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    n = (n << 3) + (n << 1) + ch - '0', ch = getchar();
  n = n * sign;
}
const int dx4[4] = {0, 0, 1, -1};
const int dy4[4] = {1, -1, 0, 0};
const int dx8[8] = {0, 0, 1, 1, 1, -1, -1, -1};
const int dy8[8] = {1, -1, 0, 1, -1, 1, 0, -1};
class node {
 public:
  int x, y, z;
  node(int a, int b, int c) {
    x = a;
    y = b;
    z = c;
  }
};
bool operator<(const node &a, const node &b) { return a.x > b.x; }
bool compare(const node &a, const node &b) { return a.x < b.x; }
int n, k;
void solve() {
  if (k > n || k == 1 && n > 1)
    cout << -1 << endl;
  else {
    if (n == 1 && k == 1)
      cout << "a\n";
    else {
      for (int i = 1; i <= n - k + 2; i++) {
        if (i & 1)
          cout << "a";
        else
          cout << "b";
      }
      for (int i = 1; i <= k - 2; i++) {
        cout << char('b' + i);
      }
      cout << endl;
    }
  }
}
void input() { cin >> n >> k; }
int main() {
  int testcase = 1;
  for (int i = 0; i < testcase; i++) {
    input();
    solve();
  }
  return 0;
}

```