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:
 
    #pragma GCC optimize ("-O3")
    #include <bits/stdc++.h>  
     
    using namespace std;
     
    typedef long long ll;
    typedef long double ld;
    typedef pair<int,int> p32;
    typedef pair<ll,ll> p64;
    typedef pair<double,double> pdd;
    typedef vector<ll> v64;
    typedef vector<int> v32;
    typedef vector<vector<int> > vv32;
    typedef vector<vector<ll> > vv64;
    typedef vector<vector<p64> > vvp64;
    typedef vector<p64> vp64;
    typedef vector<p32> vp32;
    ll MOD = 998244353;
    #define forn(i,e) for(ll i = 0; i < e; i++)
    #define forsn(i,s,e) for(int i = s; i < e; i++)
    #define rforn(i,s) for(ll i = s; i >= 0; i--)
    #define rforsn(i,s,e) for(ll i = s; i >= e; i--)
    #define ln "\n"
    #define dbg(x) cout<<#x<<" = "<<x<<ln
    #define mp make_pair
    #define pb push_back
    #define fi first
    #define se second
    #define INF 1e18
    #define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
    #define all(x) (x).begin(), (x).end()
    #define sz(x) ((ll)(x).size())
     
    ll mpow(ll a, ll b){
        if(b >= (MOD-1)) b %= (MOD-1);
        if(b==0) return 1;
        ll t1 = mpow(a,b/2);
        t1 *= t1;
        t1 %= MOD;
        if(b%2) t1 *= a;
        t1 %= MOD;
        return t1;
    }
     
    ll mpow(ll a, ll b, ll p){
        if(b==0) return 1;
        ll t1 = mpow(a,b/2,p);   t1 *= t1;
        t1 %= p;
        if(b%2) t1 *= a;
        t1 %= p;
        return t1;
    }
    ll modinverse(ll a, ll m){
        ll m0 = m; 
        ll y = 0, x = 1; 
        if (m == 1) return 0; 
        while (a > 1){
            ll q = a / m; 
            ll t = m; 
            m = a % m, a = t; 
            t = y; 
            y = x - q * y; 
            x = t; 
        }
        if (x < 0) x += m0; 
        return x; 
    }
    ll range(ll l, ll r){
        return l + rand()%(r-l+1);
    }
    
    int main()
    {
        ll t=1;
        //cin>>t;
        while(t--)
        {
            ll n,k;
            cin>>n>>k;
            string ans = "";

            for(ll i=0;i<k-1;i++)
            {
                ans += char('a'+i);
                for(ll j=i+1;j<k;j++)
                {
                    ans += char('a'+i);
                    ans += char('a' + j);
                }
            } 
            ans += char('a'+k-1);
            //ans += char('a'+ 0);
            string temp = ans;
            while(ans.length()<n)
            {
                //reverse(temp.begin(),temp.end());
                ans+=temp;
            }
            cout<<ans.substr(0,n)<<endl;
            // else
            // {
            //     for(ll i=ans.length()+1;i<=n;i++)ans+='a';
            //     cout<<ans<<endl;
            // }
        }
    }