Stepan had a favorite string s which consisted of the lowercase letters of the Latin alphabet. 

After graduation, he decided to remember it, but it was a long time ago, so he can't now remember it. But Stepan remembers some information about the string, namely the sequence of integers c_1, c_2, ..., c_{n}, where n equals the length of the string s, and c_{i} equals the number of substrings in the string s with the length i, consisting of the same letters. The substring is a sequence of consecutive characters in the string s.

For example, if the Stepan's favorite string is equal to "tttesst", the sequence c looks like: c = [7, 3, 1, 0, 0, 0, 0].

Stepan asks you to help to repair his favorite string s according to the given sequence c_1, c_2, ..., c_{n}. 


-----Input-----

The first line contains the integer n (1 ≤ n ≤ 2000) — the length of the Stepan's favorite string.

The second line contains the sequence of integers c_1, c_2, ..., c_{n} (0 ≤ c_{i} ≤ 2000), where c_{i} equals the number of substrings of the string s with the length i, consisting of the same letters.

It is guaranteed that the input data is such that the answer always exists.


-----Output-----

Print the repaired Stepan's favorite string. If there are several answers, it is allowed to print any of them. The string should contain only lowercase letters of the English alphabet. 


-----Examples-----
Input
6
6 3 1 0 0 0

Output
kkrrrq
Input
4
4 0 0 0

Output
abcd



-----Note-----

In the first test Stepan's favorite string, for example, can be the string "kkrrrq", because it contains 6 substrings with the length 1, consisting of identical letters (they begin in positions 1, 2, 3, 4, 5 and 6), 3 substrings with the length 2, consisting of identical letters (they begin in positions 1, 3 and 4), and 1 substring with the length 3, consisting of identical letters (it begins in the position 3).

Difficulty: interview

Starter code:

Solutions:
import sys

n = int(input())
c = list(map(int, input().split()))
cc = ord('a')
ans = ""
cur = n - 1
if 1:
    cur = n - 1
    while cur >= 0:
        while c[cur] > 0:
            if chr(cc) > 'z':
                cc = ord('a')
            ans += chr(cc) * (cur + 1)
            c[cur] -= 1
            for i in range(cur):
                c[i] -= (cur - i + 1)
            cc += 1
        cur -= 1
    print(ans)