#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define fast_io ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define all(v) v.begin(),v.end()

const int mod = 1e9 + 7;

int main (){
    fast_io;
    int n, k;
    cin >> n >> k;
    string ans;
    for (int i = 0; i < k; i++) {
        for (int j = i; j < k; j++) {
            ans.push_back((char)(j + 97));
            if (j + 1 != k)
                ans.push_back((char)(i + 97));
        }
    }
    int len = ans.length();
    for (int i = 0; i < n; i++) {
        cout << ans[(i % len)];
    }
    cout << endl;
}

### Prompt

Create a solution in CPP for 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;
#define ll long long
#define fast_io ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define all(v) v.begin(),v.end()

const int mod = 1e9 + 7;

int main (){
    fast_io;
    int n, k;
    cin >> n >> k;
    string ans;
    for (int i = 0; i < k; i++) {
        for (int j = i; j < k; j++) {
            ans.push_back((char)(j + 97));
            if (j + 1 != k)
                ans.push_back((char)(i + 97));
        }
    }
    int len = ans.length();
    for (int i = 0; i < n; i++) {
        cout << ans[(i % len)];
    }
    cout << endl;
}

```