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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
 
void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';} 
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
 
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
 
#define sz(x) (int)(x).size()
#define all(x) x.begin(), x.end()
typedef long long ll;

vector<int> out, euler;

void dfs(int node){
    while (out[node]){
        int next = --out[node];
        dfs(next);
    }
    euler.push_back(node);
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n, k;
    cin>>n>>k;

    out.assign(k, k);
    dfs(0);
    reverse(all(euler));
    euler.pop_back();
    string ans;
    for (int i=0;i<n;++i){
        ans += ('a'+euler[i%sz(euler)]);
    }
    cout<<ans<<"\n";

    /* int strlen, n;
    cin>>strlen>>n;
    vector<vector<int>> g(n, vector<int>(n, 1));

    vector<int> deg(n);
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j)
            deg[i] += g[i][j];
    }

    int first = 0;
    stack<int> st;
    st.push(first);
    vector<int> res;
    while (!st.empty()) {
        int v = st.top();
        int i;
        for (i = 0; i < n; ++i)
            if (g[v][i])
                break;
        if (i == n) {
            res.push_back(v);
            st.pop();
        } else {
            --g[v][i];
            st.push(i);
        }
    }
    res.pop_back();
    string ans;
    for (int i=0; i<strlen; ++i){
        ans += ('a' + res[i%(sz(res))]);
    }
    cout<<ans<<"\n"; */
    
    return 0;
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
	* ASK RELEVANT QUESTION
*/