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

const int maxn = 2e5+8, inf = 1e18+9, mod = 1e9+7;

char s[maxn];
int n, k;

void solve() {
    int i, j, idx = 0;
    cin >> n >> k; s[n + 1] = 0;
#define check() do { if (idx == n) return (void)(cout << (s + 1) << endl); } while (0)
    while (idx <= n) {
        for (i = 0; i < k; i++) {
            s[++idx] = 'a' + i; check();
            for (j = i + 1; j < k; j++) {
                s[++idx] = 'a' + i; check();
                s[++idx] = 'a' + j; check();
            }
        }
    }
}

signed main() {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0); cout << fixed << setprecision(15);
    int t = 1; //cin >> t;
    while (t--) solve();
    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>
#define int long long
using namespace std;

const int maxn = 2e5+8, inf = 1e18+9, mod = 1e9+7;

char s[maxn];
int n, k;

void solve() {
    int i, j, idx = 0;
    cin >> n >> k; s[n + 1] = 0;
#define check() do { if (idx == n) return (void)(cout << (s + 1) << endl); } while (0)
    while (idx <= n) {
        for (i = 0; i < k; i++) {
            s[++idx] = 'a' + i; check();
            for (j = i + 1; j < k; j++) {
                s[++idx] = 'a' + i; check();
                s[++idx] = 'a' + j; check();
            }
        }
    }
}

signed main() {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0); cout << fixed << setprecision(15);
    int t = 1; //cin >> t;
    while (t--) solve();
    return 0;
}


```