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

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

#define NFS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define fr(i,a,b) for(ll i=a; i<b; i++)
#define rf(i,a,b) for(ll i=a; i>=b; i--)
#define endl "\n"

#define err() cout<<"=================================="<<endl;
#define errA(A) for(auto i:A)   cout<<i<<" ";cout<<endl;
#define err1(a) cout<<#a<<" "<<a<<endl
#define err2(a,b) cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<endl
#define err3(a,b,c) cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<" "<<#c<<" "<<c<<endl
#define err4(a,b,c,d) cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<" "<<#c<<" "<<c<<" "<<#d<<" "<<d<<endl

#define ll long long int 
#define ull unsigned long long 
#define vll vector<long long>
#define mll map<long long,long long>
#define pll pair<long long,long long>
#define v2d vector<vll>
#define vpll vector<pll>
#define mpll map<pll>
#define all(v) v.begin(),v.end()
#define mx(a) *max_element(all(a))
#define mn(a) *min_element(all(a))
#define alla(v) v,v+n
#define acc(a) accumulate(alla(a),(ll)0)
#define PB push_back
#define MP make_pair
#define mod 1000000007
#define PI 3.14159265358979323846


ll mul(ll a,ll b) {
	return ((a % mod) * (b % mod)) % mod;
}
 
ll add(ll a,ll b) {
	return (a % mod + b % mod) % mod;
}
ll gcd(ll a, ll b){
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
ll extgcd(ll a,ll b,ll& x,ll& y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    ll x1, y1;
    ll d = extgcd(b, a % b, x1, y1);
    x = y1;
    y = x1 - y1 * (a / b);
    return d;
}
bool isPrime(ll n){
    if (n <= 1) 
        return false; 
    if (n <= 3) 
        return true; 
    if (n % 2 == 0 || n % 3 == 0) 
        return false; 
    for (ll i = 5; i * i <= n; i = i + 6) 
        if (n % i == 0 || n % (i + 2) == 0) 
            return false; 
    return true; 
}
bool isPowerOfTwo(ll n) { 
   if(n==0) 
   return false; 
   return (ceil(log2(n)) == floor(log2(n))); 
}
//Prime factorization of integer n O(sqrt(n))
vector<ll> factors(ll  n)
{
    vector<ll> f;
    for (ll x = 2; x*x <= n; x++)
    {
        while (n%x == 0)
        {
            f.push_back(x);
            n /= x;
        }
    }
    if (n > 1) f.push_back(n);
    return f;
}
ll bin_exp_mod(ll a,ll n)
{
    ll res=1;
    a%=mod;
    if(a==0)
        return 0;
    while(n>0)
    {
        if(n&1)
            res=((res%mod)*(a%mod))%mod;
        n=n>>1;
        a=((a%mod)*(a%mod))%mod;
    }
    res%=mod;
    return res;
}

void solve()
{
    ll n,m,k;
    cin>>n>>k;
    
    string str ="";
    char a = 'a',b;
    fr(i,0,k)
    {
        str +=a;
        b=a;
        b++;
        fr(j,i,k)
        {
            if(j == k-1)
                continue;
            str+=a;
            str+=b;
            b++;
        }
        a++;
    }
    ll n1 = str.length();
    //cout<<str<<"\n";
    
    fr(i,0,n)
    {
        cout<<str[(i)%n1];
    }
    
    cout<<"\n";
}
int main() {
	// your code goes here
	NFS;
	    solve();
	return 0;
}

            