CodeForces Round #527 (Div3) A. Uniform String

http://codeforces.com/contest/1092/problem/A

You are given two integers nn and kk.

Your task is to construct such a string ss of length nn that for each ii from 11 to kk there is at least one ii-th letter of the Latin alphabet in this string (the first letter is 'a', the second is 'b' and so on) and there are no other letters except these. You have to maximize the minimal frequency of some letter (the frequency of a letter is the number of occurrences of this letter in a string). If there are several possible answers, you can print any.

You have to answer tt independent queries.

Input

The first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of queries.

The next tt lines are contain queries, one per line. The ii-th line contains two integers nini and kiki (1≤ni≤100,1≤ki≤min(ni,26)1≤ni≤100,1≤ki≤min(ni,26)) — the length of the string in the ii-th query and the number of characters in the ii-th query.

Output

Print tt lines. In the ii-th line print the answer to the ii-th query: any string sisi satisfying the conditions in the problem statement with constraints from the ii-th query.

Example

input

Copy

3
7 3
4 4
6 2

output

Copy

cbcacab
abcd
baabab

Note

In the first example query the maximum possible minimal frequency is 22, it can be easily seen that the better answer doesn't exist. Other examples of correct answers: "cbcabba", "ccbbaaa" (any permutation of given answers is also correct).

In the second example query any permutation of first four letters is acceptable (the maximum minimal frequency is 11).

In the third example query any permutation of the given answer is acceptable (the maximum minimal frequency is 33).

代码：

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

int T;

int main() {
    scanf("%d", &T);
    while(T --) {
        int N, K;
        scanf("%d%d", &N, &K);
        string s = "";
        if(K > N) {
            for(int i = 0; i < N; i ++)
                s += ('a' + i);
        } else {
            int cnt = N / K;
            for(int i = 0; i < cnt; i ++) {
                for(int j = 0; j < K; j ++)
                    s += ('a' + j);
            }

            if(N % K) {
                N = N - K * cnt;
                for(int i = 0; i < N; i ++)
                    s += 'a';
            }
        }

        cout << s << endl;
    }
    return 0;
}

　　被期末论文大作业支配的一周！！！好久没写题了