1511_D. Min Cost String
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
```python

def createks(k):
    ks = ['a', 'a']
    if k == 1:
        return ks
    ks = ['a', 'a', 'b', 'b', 'a']
    if k == 2:
        return ks
    # k >= 3, follows formula.
    msd = 2
    while msd < k:
        #create extension with msd.
        ks.extend([chr(ord('a') + msd), chr(ord('a') + msd-1), chr(ord('a') + msd)])
        for i in range(msd-1):
            ks.extend(chr(ord('a') + msd))
            ks.extend(chr(ord('a') + (msd-(2+i))))
        msd += 1
    #print('ks=', ks)
    return ks #ks will be a list of chars, rather than a string.


def answer(n, k):
    ks = createks(k)
    ans = []
    extra = n - len(ks)
    for i in range(min(n, len(ks))):
            ans.append(ks[i]) #cost = 0
    
    ks.pop(0) #n thing to do. But only done once.
    for i in range(extra):
        ans.append(ks[i%len(ks)])

    return ''.join(ans)

def main():
    n, k = [int(i) for i in input().split()] # 1 <=n <= 2*10**5, 1 <= k <= 26
    print(answer(n, k))

    return
main()
```