#include<bits/stdc++.h>
using namespace std;

char enc(int k) {
    return 'a' + k;
}

string solve(int k) {
    string res;
    char c = enc(k);
    if (k == 0) {
        res.push_back(c);
        return res;
    }
    res.push_back(c);
    res.push_back(c);
    for (int i = 0; i < k - 1; i++) {
        res.push_back(enc(i));
        res.push_back(c);
    }
    res += solve(k - 1);
    res.push_back(enc(k - 1));
    return res;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, k;
    cin >> n >> k;
    string s = solve(k - 1);
    // cout << s << endl;

    string res;
    for (int i = 0; i < n; i++) {
        res.push_back(s[i % s.size()]);
    }
    cout << res << '\n';

    return 0;
}

### Prompt

Generate a CPP solution to the following problem:

Let's define the cost of a string s as the number of index pairs i and j (1 ≤ i < j < |s|) such that s_i = s_j and s_{i+1} = s_{j+1}.

You are given two positive integers n and k. Among all strings with length n that contain only the first k characters of the Latin alphabet, find a string with minimum possible cost. If there are multiple such strings with minimum cost — find any of them.

Input

The only line contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 26).

Output

Print the string s such that it consists of n characters, each its character is one of the k first Latin letters, and it has the minimum possible cost among all these strings. If there are multiple such strings — print any of them.

Examples

Input


9 4


Output


aabacadbb


Input


5 1


Output


aaaaa

Input


10 26


Output


codeforces

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

char enc(int k) {
    return 'a' + k;
}

string solve(int k) {
    string res;
    char c = enc(k);
    if (k == 0) {
        res.push_back(c);
        return res;
    }
    res.push_back(c);
    res.push_back(c);
    for (int i = 0; i < k - 1; i++) {
        res.push_back(enc(i));
        res.push_back(c);
    }
    res += solve(k - 1);
    res.push_back(enc(k - 1));
    return res;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, k;
    cin >> n >> k;
    string s = solve(k - 1);
    // cout << s << endl;

    string res;
    for (int i = 0; i < n; i++) {
        res.push_back(s[i % s.size()]);
    }
    cout << res << '\n';

    return 0;
}

```