#include<bits/stdc++.h>

#define int long long
#define fi first
#define se second

using namespace std;

//const int MAXN = 200005;
//const int MOD = 1e9 + 7;
//const int INF = 1e18;

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios_base :: sync_with_stdio(0);
    int n, k;
    cin >> n >> k;
    char mx = 'a' + k - 1;
    set<char> st[150];
    for (char c1 = 'a'; c1 <= mx; c1++) {
        for (char c2 = 'a'; c2 <= mx; c2++) {
            st[c1].insert(c2);
        }
    }
    string s = "";
    char x = mx;
    while (!st[x].empty()) {
        s += x;
        int to = *st[x].begin();
        st[x].erase(to);
        x = to;
    }
    for (int i = 0; i < n; i++) {
        cout << s[i % s.size()];
    }
    return 0;
}

### 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>

#define int long long
#define fi first
#define se second

using namespace std;

//const int MAXN = 200005;
//const int MOD = 1e9 + 7;
//const int INF = 1e18;

signed main() {
    cin.tie(0);
    cout.tie(0);
    ios_base :: sync_with_stdio(0);
    int n, k;
    cin >> n >> k;
    char mx = 'a' + k - 1;
    set<char> st[150];
    for (char c1 = 'a'; c1 <= mx; c1++) {
        for (char c2 = 'a'; c2 <= mx; c2++) {
            st[c1].insert(c2);
        }
    }
    string s = "";
    char x = mx;
    while (!st[x].empty()) {
        s += x;
        int to = *st[x].begin();
        st[x].erase(to);
        x = to;
    }
    for (int i = 0; i < n; i++) {
        cout << s[i % s.size()];
    }
    return 0;
}

```