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>
//#pragma GCC optimize ("03")
#define FastIO ios_base::sync_with_stdio(false) , cin.tie(0) , cout.tie(0)
#define FILES freopen("in" , "r" , stdin) , freopen("out" , "w" , stdout)
#define ll long long
#define ull unsigned long long
#define ld long double
#define eb emplace_back
#define pb push_back
#define qwerty1 first
#define qwerty2 second
#define qwerty3 -> first
#define qwerty4 -> second
#define umap unordered_map
#define uset unordered_set
#define pii pair < int , int >
#define pq priority_queue
#define dbg(x) cerr << #x << ": " << x << '\n'

namespace FastRead
{
    char __buff[5000];int __lg = 0 , __p = 0;
    char nc()
    {
        if(__lg == __p){__lg = fread(__buff , 1 , 5000 , stdin);__p = 0;if(!__lg) return EOF;}
        return __buff[__p++];
    }
    template<class T>void read(T&__x)
    {
        T __sgn = 1; char __c;while(!isdigit(__c = nc()))if(__c == '-')__sgn = -1;
        __x = __c - '0';while(isdigit(__c = nc()))__x = __x * 10 + __c - '0';__x *= __sgn;
    }
}

using namespace FastRead;
using namespace std;

const int N = 1e5 + 10;
const int M = 1e9 + 7;
const ld PI = acos(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n , k;
vector < int > G[30];
string a;

signed main()
{
	#ifndef ONLINE_JUDGE
		FastIO , FILES;
	#endif

    int i , j;

    cin >> n >> k;

    for(i = 1 ; i <= k ; i++)
        for(j = 1 ; j <= k ; j++)
            G[i].pb(j);

    stack < int > s;
    s.push(1);

    while(!s.empty())
    {
        int w = s.top();

        if(G[w].empty())
        {
            a += (char)(w - 1 + 'a');
            s.pop();
        }
        else
        {
            s.push(G[w].back());
            G[w].pop_back();
        }
    }

    a.pop_back();

    while(a.size() < n)
        a += a;

    cout << a.substr(0 , n);

    return 0;
}